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Simplify. 1. i 10 SOLUTION: 2. i 2 + i 8 SOLUTION: 3. i 3 + i 20 SOLUTION: 4. i 100 SOLUTION: 5. i 77 SOLUTION: 4 12 eSolutions Manual - Powered by Cognero Page 1 0-2 Operations with Complex Numbers

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Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 1

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 2

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 3

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 4

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 5

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 6

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 7

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 8

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 9

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 10

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 11

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 12

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 13

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 14

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 15

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

eSolutions Manual - Powered by Cognero Page 16

0-2 Operations with Complex Numbers

Simplify.

1. i–10

SOLUTION:  

2. i2 + i8

SOLUTION:  

3. i3 + i20

SOLUTION:  

4. i100

SOLUTION:  

5. i77

SOLUTION:  

6. i4 + i–12

SOLUTION:  

7. i5 + i9

SOLUTION:  

8. i18

SOLUTION:  

Simplify.9. (3 + 2i) + (–4 + 6i)

SOLUTION:  

10. (7 – 4i) + (2 – 3i)

SOLUTION:  

11. (0.5 + i) – (2 – i)

SOLUTION:  

12. (–3 – i) – (4 – 5i)

SOLUTION:  

13. (2 + 4.1i) – (–1 – 6.3i)

SOLUTION:  

14. (2 + 3i) + (–6 + i)

SOLUTION:  

15. (–2 + 4i) + (5 – 4i)

SOLUTION:  

16. (5 + 7i) – (–5 + i)

SOLUTION:  

17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I asa variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit.

SOLUTION:  

Simplify.

18. (–2 – i)2

SOLUTION:  

19. (1 + 4i)2

SOLUTION:  

20. (5 + 2i)2

SOLUTION:  

21. (3 + i)2

SOLUTION:  

22. (2 + i)(4 + 3i)

SOLUTION:  

23. (3 + 5i)(3 – 5i)

SOLUTION:  

24. (5 + 3i)(2 + 6i)

SOLUTION:  

25. (6 + 7i)(6 – 7i)

SOLUTION:  

Simplify.

26. 

SOLUTION:  

27. 

SOLUTION:  

28. 

SOLUTION:  

29. 

SOLUTION:  

30. 

SOLUTION:  

31. 

SOLUTION:  

32. 

SOLUTION:  

33. 

SOLUTION:  

ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance.

34. I = 1 + 3j amps, Z = 7 – 5j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 22 + 16j volts.

35. I = 2 – 7j amps, Z = 4 – 3j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is –13 – 34j volts.

36. I = 5 – 4j amps, Z = 3 + 2j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 23 – 2j volts.

37. I = 3 + 10j amps, Z = 6 – j ohms

SOLUTION:  Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit.

Therefore, the voltage of the circuit is 28 + 57j volts.

Solve each equation.

38. 5x2 + 5 = 0

SOLUTION:  

39. 4x2 + 64 = 0

SOLUTION:  

40. 2x2 + 12 = 0

SOLUTION:  

41. 6x2 + 72 = 0

SOLUTION:  

42. 8x2 + 120 = 0

SOLUTION:  

43. 3x2 + 507 = 0

SOLUTION:  

44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance ,

and the reactance due to inductance , and can be written as a complex number R + (XL  – XC)j . The values (in

ohms) for R, , and in the first and second parts of a particular series circuit are shown.

a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms.

SOLUTION:  

a. Substitute the given values for R,  XC, and  into the equation Z = R + (XL  – XC)j .

The impedance in the first part of the circuit is  ohms and in the second of the circuit is   ohms. b.  ohms

c. The reciprocal of a number x is , so the reciprocal of  is  . Simplify this fraction by rationalizing the

denominator.

 

Find the values of x and y to make each equation true.

45. 3x + 2iy = 6 + 10i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

46. 5x + 3iy = 5 – 6i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

47. x – iy = 3 + 4i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

48. –5x + 3iy = 10 – 9i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

49. 2x + 3iy = 12 + 12i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

50. 4x – iy = 8 + 7i

SOLUTION:  To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the

equation  and then solve for x and y respectively.

Simplify.51. (2 – i)(3 + 2i)(1 – 4i)

SOLUTION:  

52. (–1 – 3i)(2 + 2i)(1 – 2i)

SOLUTION:  

53. (2 + i)(1 + 2i)(3 – 4i)

SOLUTION:  

54. (−5 – i)(6i + 1)(7 – i)

SOLUTION:  

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0-2 Operations with Complex Numbers