complex numbers · complex numbers a complex numbercan be represented by an expression of the form...
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COMPLEX NUMBERS
A complex number can be represented by an expression of the form , where andare real numbers and is a symbol with the property that . The complex num-
ber can also be represented by the ordered pair and plotted as a point in aplane (called the Argand plane) as in Figure 1. Thus, the complex number isidentified with the point .
The real part of the complex number is the real number and the imaginarypart is the real number . Thus, the real part of is and the imaginary part is .Two complex numbers and are equal if and , that is, their realparts are equal and their imaginary parts are equal. In the Argand plane the horizontal axisis called the real axis and the vertical axis is called the imaginary axis.
The sum and difference of two complex numbers are defined by adding or subtractingtheir real parts and their imaginary parts:
For instance,
The product of complex numbers is defined so that the usual commutative and distributivelaws hold:
Since , this becomes
EXAMPLE 1
EXAMPLE 2 Express the number in the form .
SOLUTION We multiply numerator and denominator by the complex conjugate of ,namely , and we take advantage of the result of Example 1:
The geometric interpretation of the complex conjugate is shown in Figure 2: is thereflection of in the real axis. We list some of the properties of the complex conjugate inthe following box. The proofs follow from the definition and are requested in Exercise 18.
Properties of Conjugates
z n � z nzw � z wz � w � z � w
zz
�1 � 3i
2 � 5i�
�1 � 3i
2 � 5i�
2 � 5i
2 � 5i�
13 � 11i
22 � 52 �13
29�
11
29 i
2 � 5i2 � 5i
a � bi�1 � 3i
2 � 5i
� �2 � 5i � 6i � 15��1� � 13 � 11i
��1 � 3i ��2 � 5i � � ��1��2 � 5i � � 3i�2 � 5i �
�a � bi ��c � di � � �ac � bd � � �ad � bc�i
i 2 � �1
� ac � adi � bci � bdi 2
�a � bi ��c � di � � a�c � di � � �bi ��c � di �
�1 � i � � �4 � 7i � � �1 � 4� � ��1 � 7�i � 5 � 6i
�a � bi � � �c � di � � �a � c� � �b � d �i
�a � bi � � �c � di � � �a � c� � �b � d �i
b � da � cc � dia � bi�344 � 3ib
aa � bi�0, 1�
i � 0 � 1 � i�a, b�a � bi
i 2 � �1ibaa � bi
1
FIGURE 1Complex numbers as points inthe Argand plane
Re
Im
0
i
_2-2i
_i
3-2i
2+3i
_4+2i
1
Re
Im
0
i
_i
z=a-bi–
z=a+bi
FIGURE 2
Division of complex numbers is much like rationalizing the denominator of a rationalexpression. For the complex number , we define its complex conjugate to be
. To find the quotient of two complex numbers we multiply numerator anddenominator by the complex conjugate of the denominator.z � a � bi
z � a � bi
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The modulus, or absolute value, of a complex number is its distancefrom the origin. From Figure 3 we see that if , then
Notice that
and so
This explains why the division procedure in Example 2 works in general:
Since , we can think of as a square root of . But notice that we also haveand so is also a square root of . We say that is the principal
square root of and write . In general, if is any positive number, we write
With this convention, the usual derivation and formula for the roots of the quadratic equa-tion are valid even when :
EXAMPLE 3 Find the roots of the equation .
SOLUTION Using the quadratic formula, we have
We observe that the solutions of the equation in Example 3 are complex conjugates ofeach other. In general, the solutions of any quadratic equation with realcoefficients , , and are always complex conjugates. (If is real, , so is its ownconjugate.)
We have seen that if we allow complex numbers as solutions, then every quadratic equa-tion has a solution. More generally, it is true that every polynomial equation
of degree at least one has a solution among the complex numbers. This fact is known asthe Fundamental Theorem of Algebra and was proved by Gauss.
POLAR FORM
We know that any complex number can be considered as a point and thatany such point can be represented by polar coordinates with . In fact,
as in Figure 4. Therefore, we have
z � a � bi � �r cos �� � �r sin ��i
b � r sin �a � r cos �
r � 0�r, ���a, b�z � a � bi
anxn � an�1xn�1 � � � � � a1 x � a0 � 0
zz � zzcbaax 2 � bx � c � 0
x ��1 � s1 2 � 4 � 1
2�
�1 � s�3
2�
�1 � s3 i
2
x 2 � x � 1 � 0
x ��b � sb 2 � 4ac
2a
b 2 � 4ac � 0ax 2 � bx � c � 0
s�c � sc i
cs�1 � i�1i�1�i��i�2 � i 2 � �1
�1ii 2 � �1
z
w�
zw
ww�
zw
� w �2
zz � � z �2
zz � �a � bi��a � bi� � a 2 � abi � abi � b 2i 2 � a 2 � b 2
� z � � sa 2 � b 2
z � a � biz � a � bi� z �
Re
Im
0
a+bi
b
¨
r
a
FIGURE 4
2 ■ COMPLEX NUMBERS
FIGURE 3
Re
Im
0
bi
a
b
z=a+bi
|z |= a
@+b@
œ„„„„„
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Thus, we can write any complex number in the form
where
The angle is called the argument of and we write . Note that is notunique; any two arguments of differ by an integer multiple of .
EXAMPLE 4 Write the following numbers in polar form.
(a) (b)
SOLUTION(a) We have and , so we can take .Therefore, the polar form is
(b) Here we have and . Since lies in thefourth quadrant, we take and
The numbers and are shown in Figure 5.
The polar form of complex numbers gives insight into multiplication and division. Let
be two complex numbers written in polar form. Then
Therefore, using the addition formulas for cosine and sine, we have
This formula says that to multiply two complex numbers we multiply the moduli and addthe arguments. (See Figure 6.)
A similar argument using the subtraction formulas for sine and cosine shows that todivide two complex numbers we divide the moduli and subtract the arguments.
In particular, taking and , (and therefore and ), we have the fol-lowing, which is illustrated in Figure 7.
1
z�
1
r �cos � � i sin ��.thenz � r�cos � � i sin ��,If
� 2 � ��1 � 0z2 � zz1 � 1
z2 � 0z1
z2�
r1
r2 �cos��1 � �2 � � i sin��1 � �2 ��
z1z2 � r1r2�cos��1 � �2 � � i sin��1 � �2 ��1
� r1r2��cos �1 cos �2 � sin �1 sin �2 � � i�sin �1 cos �2 � cos �1 sin �2 ��
z1z2 � r1r2�cos �1 � i sin �1��cos �2 � i sin �2 �
z2 � r2�cos �2 � i sin �2 �z1 � r1�cos �1 � i sin �1�
wz
w � 2�cos���
6 � � i sin���
6 �� � ��6
wtan � � �1s3 r � � w � � s3 � 1 � 2
z � s2 �cos �
4� i sin
�
4 �� � �4tan � � 1r � � z � � s12 � 12 � s2
w � s3 � iz � 1 � i
2�zarg�z�� � arg�z�z�
tan � �b
aandr � � z � � sa 2 � b 2
z � r�cos � � i sin ��
z
Re
Im
0
œ„3-i
2
1+i
œ„2
π4
_π6
FIGURE 5
z¡
FIGURE 6
Re
Im
z¡z™
¨¡+¨™
z™
¨¡
¨™
Re
Im
0
r
z
¨
_¨
1r
1z
FIGURE 7
COMPLEX NUMBERS ■ 3
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EXAMPLE 5 Find the product of the complex numbers and in polar form.
SOLUTION From Example 4 we have
and
So, by Equation 1,
This is illustrated in Figure 8.
Repeated use of Formula 1 shows how to compute powers of a complex number. If
then
and
In general, we obtain the following result, which is named after the French mathematicianAbraham De Moivre (1667–1754).
De Moivre’s Theorem If and is a positive integer, then
This says that to take the nth power of a complex number we take the nth power of themodulus and multiply the argument by n.
EXAMPLE 6 Find .
SOLUTION Since , it follows from Example 4(a) that has the polarform
So by De Moivre’s Theorem,
De Moivre’s Theorem can also be used to find the th roots of complex numbers. An th root of the complex number is a complex number such that
wn � z
wznn
�25
210 �cos 5�
2� i sin
5�
2 � �1
32 i
�1
2�
1
2 i�10
� �s2
2 �10�cos 10�
4� i sin
10�
4 �
1
2�
1
2 i �
s2
2 �cos
�
4� i sin
�
4 �12 �
12 i1
2 �12 i � 1
2 �1 � i�
( 12 �
12 i)10
z n � �r �cos � � i sin ���n � r n�cos n� � i sin n��
nz � r �cos � � i sin ��2
z 3 � zz 2 � r 3�cos 3� � i sin 3��
z 2 � r 2�cos 2� � i sin 2��
z � r �cos � � i sin ��
� 2s2 �cos �
12� i sin
�
12� �1 � i�(s3 � i) � 2s2 �cos��
4�
�
6 � � i sin��
4�
�
6 �
s3 � i � 2�cos���
6 � � i sin���
6 � 1 � i � s2 �cos
�
4� i sin
�
4 �
s3 � i1 � i
4 ■ COMPLEX NUMBERS
0
2
z=1+i
w=œ„3-i
zw2œ„2œ„2
FIGURE 8
Re
Im
π12
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Writing these two numbers in trigonometric form as
and using De Moivre’s Theorem, we get
The equality of these two complex numbers shows that
and
From the fact that sine and cosine have period it follows that
Thus
Since this expression gives a different value of for , 1, 2, . . . , , we have thefollowing.
Roots of a Complex Number Let and let be a positive inte-ger. Then has the distinct th roots
where , 1, 2, . . . , .
Notice that each of the th roots of has modulus . Thus, all the th rootsof lie on the circle of radius in the complex plane. Also, since the argument of eachsuccessive th root exceeds the argument of the previous root by , we see that the th roots of are equally spaced on this circle.
EXAMPLE 7 Find the six sixth roots of and graph these roots in the complexplane.
SOLUTION In trigonometric form, . Applying Equation 3 with ,we get
We get the six sixth roots of by taking in this formula:
w2 � 816�cos 5�
6� i sin
5�
6 � � s2 ��s3
2�
1
2 i�
w1 � 816�cos �
2� i sin
�
2 � � s2 i
w0 � 816�cos �
6� i sin
�
6 � � s2 �s3
2�
1
2 i�
k � 0, 1, 2, 3, 4, 5�8
wk � 816�cos � � 2k�
6� i sin
� � 2k�
6 �n � 6z � 8�cos � � i sin ��
z � �8
zn2�nn
r 1nzn� wk � � r 1nzn
n � 1k � 0
wk � r 1n�cos� � � 2k�
n � � i sin� � � 2k�
n �nnz
nz � r �cos � � i sin ��3
n � 1k � 0w
w � r 1n�cos� � � 2k�
n � � i sin� � � 2k�
n � �
� � 2k�
norn � � � 2k�
2�
sin n � sin �andcos n � cos �
s � r 1norsn � r
sn�cos n � i sin n� � r �cos � � i sin ��
z � r �cos � � i sin ��andw � s�cos � i sin �
COMPLEX NUMBERS ■ 5
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6 ■ COMPLEX NUMBERS
All these points lie on the circle of radius as shown in Figure 9.
COMPLEX EXPONENTIALS
We also need to give a meaning to the expression when is a complex num-ber. The theory of infinite series as developed in Chapter 8 can be extended to the casewhere the terms are complex numbers. Using the Taylor series for we define
and it turns out that this complex exponential function has the same properties as the realexponential function. In particular, it is true that
If we put , where is a real number, in Equation 4, and use the facts that
. . .
we get
Here we have used the Taylor series for and (Equations 8.7.17 and 8.7.16). The result is a famous formula called Euler’s formula:
Combining Euler’s formula with Equation 5, we get
ex�iy � exe iy � ex�cos y � i sin y�7
e iy � cos y � i sin y6
sin ycos y
� cos y � i sin y
� �1 � y 2
2!�
y 4
4!�
y 6
6!� � � �� � i�y �
y 3
3!�
y 5
5!� � � ��
� 1 � iy � y 2
2!� i
y 3
3!�
y 4
4!� i
y 5
5!� � � �
e iy � 1 � iy ��iy�2
2!�
�iy�3
3!�
�iy�4
4!�
�iy�5
5!� � � �
i 5 � i,i 4 � 1,i 3 � i 2i � �i,i 2 � �1,
yz � iy
e z1�z2 � e z1e z25
e z � �
n�0 z n
n!� 1 � z �
z2
2!�
z3
3!� � � �4
ex
z � x � iye z
s2
w5 � 816�cos 11�
6� i sin
11�
6 � � s2 �s3
2�
1
2 i�
w4 � 816�cos 3�
2� i sin
3�
2 � � �s2 i
w3 � 816�cos 7�
6� i sin
7�
6 � � s2 ��s3
2�
1
2 i�
FIGURE 9The six sixth roots of z=_8
0
w¡
w¢
w∞
w¸w™
w£
_œ„2 œ„2
_œ„2i
œ„2i
Re
Im
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2(8.7.12) as our guide,
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COMPLEX NUMBERS ■ 7
EXAMPLE 8 Evaluate: (a) (b)
SOLUTION(a) From Euler’s equation (6) we have
(b) Using Equation 7 we get
Finally, we note that Euler’s equation provides us with an easier method of proving De Moivre’s Theorem:
�r �cos � � i sin ���n � �re i� �n � r ne in� � r n�cos n� � i sin n��
e�1�i�2 � e�1�cos �
2� i sin
�
2 � �1
e �0 � i�1�� �
i
e
e i� � cos � � i sin � � �1 � i�0� � �1
e�1�i�2e i�
■ ■ We could write the result of Example 8(a)as
This equation relates the five most famous numbers in all of mathematics: and .�0, 1, e, i,
e i� � 1 � 0
EXERCISES
1–14 Evaluate the expression and write your answer in the form .
1. 2.
3. 4.
5. 6.
7. 8.
9. 10.
11. 12.
13. 14.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
15–17 Find the complex conjugate and the modulus of the number.
15. 16.
17.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
18. Prove the following properties of complex numbers.(a) (b)(c) , where is a positive integer
[Hint: Write , .]
19–24 Find all solutions of the equation.
19. 20.
21. 22.
23. 24.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
z2 �12 z �
14 � 0z2 � z � 2 � 0
2x 2 � 2x � 1 � 0x 2 � 2x � 5 � 0
x 4 � 14x 2 � 9 � 0
w � c � diz � a � binz n � z n
zw � z wz � w � z � w
�4i
�1 � 2s2 i12 � 5i
s�3s�12s�25
i 100i 3
3
4 � 3i
1
1 � i
3 � 2i
1 � 4i
1 � 4i
3 � 2i
2i ( 12 � i)12 � 7i
�1 � 2i ��8 � 3i ��2 � 5i ��4 � i�
(4 �12 i) � (9 �
52 i)�5 � 6i � � �3 � 2i �
a � bi
25–28 Write the number in polar form with argument between and .
25. 26.
27. 28.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
29–32 Find polar forms for , , and by first putting and into polar form.
29. ,
30. ,
31. ,
32. ,■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
33–36 Find the indicated power using De Moivre’s Theorem.
33. 34.
35. 36.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
37–40 Find the indicated roots. Sketch the roots in the complexplane.
37. The eighth roots of 1 38. The fifth roots of 32
39. The cube roots of 40. The cube roots of ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
41–46 Write the number in the form .
41. 42.
43. 44.
45. 46.■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
47. Use De Moivre’s Theorem with to express andin terms of and .sin �cos �sin 3�
cos 3�n � 3
e��ie 2�i�
e �i�e i�3
e 2� ie i�2
a � bi
1 � ii
�1 � i �8(2s3 � 2i )5
(1 � s3 i )5�1 � i �20
w � �3 � 3iz � 4(s3 � i )w � �1 � iz � 2s3 � 2i
w � 8iz � 4s3 � 4i
w � 1 � s3 iz � s3 � i
wz1zzwzw
8i3 � 4i
1 � s3 i�3 � 3i
2�0
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50. (a) If is a complex-valued function of a real variable, itsindefinite integral is an antiderivative of . Evaluate
(b) By considering the real and imaginary parts of the integralin part (a), evaluate the real integrals
and
(c) Compare with the method used in Example 4 in Section 6.1.
y e x sin x dxy e x cos x dx
y e �1�i �x dx
ux u�x� dxu
8 ■ COMPLEX NUMBERS
48. Use Euler’s formula to prove the following formulas for and :
49. If is a complex-valued function of a realvariable and the real and imaginary parts and aredifferentiable functions of , then the derivative of is definedto be . Use this together with Equation 7to prove that if , then when is a complex number.
r � a � biF��x� � re rxF�x� � e rxu��x� � f ��x� � it��x�
uxt�x�f �x�x
u�x� � f �x� � it�x�
sin x �eix � e�ix
2icos x �
eix � e�ix
2
sin xcos x
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COMPLEX NUMBERS ■ 9
ANSWERS
1. 3.
5. 7.
9. 11. 13. 15.
17. 19.
21. 23.
25.
27.
29. ,12 �cos���6� � i sin���6��
4�cos��2� � i sin��2��, cos���6� � i sin���6�
5{cos[tan�1(43)] � i sin[tan�1(4
3)]}3s2 �cos�3�4� � i sin�3�4��
�12 � (s72)i�1 � 2i
�32 i4i, 4
12 � 5i; 135i�i12 �
12 i
1113 �
1013 i12 � 7i
13 � 18i8 � 4i
31. ,,
33. 35.
37. 39.
41. 43. 45.
47. , sin 3� � 3 cos2� sin � � sin3�cos 3� � cos3� � 3 cos � sin2�
�e 212 � (s32)ii
0
Im
Re
_i
0
Im
Re
i
1
�(s32) �12 i, �i�1, �i, (1s2)��1 � i �
�512s3 � 512i�1024
14 �cos��6� � i sin��6��(2s2)�cos�13�12� � i sin�13�12��
4s2 �cos�7�12� � i sin�7�12��Click here for solutions.S
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10 ■ COMPLEX NUMBERS
SOLUTIONS
1. (5− 6i) + (3 + 2i) = (5 + 3) + (−6 + 2)i = 8 + (−4)i = 8− 4i
2. 4− 12i − 9 + 5
2i = (4− 9) + − 1
2− 5
2i = −5 + (−3)i = −5− 3i
3. (2 + 5i)(4− i) = 2(4) + 2(−i) + (5i)(4) + (5i)(−i) = 8− 2i+ 20i− 5i2= 8 + 18i− 5(−1) = 8 + 18i+ 5 = 13 + 18i
4. (1− 2i)(8− 3i) = 8− 3i− 16i+ 6(−1) = 2− 19i
5. 12 + 7i = 12− 7i
6. 2i 12− i = i− 2(−1) = 2 + i ⇒ 2i 1
2− i = 2 + i = 2− i
7.1 + 4i
3 + 2i=1 + 4i
3 + 2i· 3− 2i3− 2i =
3− 2i+ 12i− 8(−1)32 + 22
=11 + 10i
13=11
13+10
13i
8.3 + 2i
1− 4i =3 + 2i
1− 4i ·1 + 4i
1 + 4i=3+ 12i+ 2i+ 8(−1)
12 + 42=−5 + 14i17
= − 5
17+14
17i
9.1
1 + i=
1
1 + i· 1− i
1− i=
1− i
1− (−1) =1− i
2=1
2− 1
2i
10.3
4− 3i =3
4− 3i ·4 + 3i
4 + 3i=
12 + 9i
16− 9(−1) =12
25+9
25i
11. i3 = i2 · i = (−1)i = −i
12. i100 = i250= (−1)50 = 1
13.√−25 = √25 i = 5i
14.√−3√−12 = √3 i√12 i = √3 · 12 i2 = √36 (−1) = −6
15. 12− 5i = 12 + 15i and |12− 15i| = 122 + (−5)2 = √144 + 25 = √169 = 13
16. −1 + 2√2 i = −1− 2√2 i and −1 + 2√2 i = (−1)2 + 2√2
2=√1 + 8 =
√9 = 3
17. −4i = 0− 4i = 0 + 4i = 4i and |−4i| = 02 + (−4)2 = √16 = 4
18. Let z = a+ bi, w = c+ di.
(a) z + w = (a+ bi) + (c+ di) = (a+ c) + (b+ d)i
= (a+ c)− (b+ d)i = (a− bi) + (c− di) = z +w
(b) zw = (a+ bi)(c+ di) = (ac− bd) + (ad+ bc)i = (ac− bd)− (ad+ bc)i.
On the other hand, z w = (a− bi)(c− di) = (ac− bd)− (ad+ bc)i = zw.
(c) Use mathematical induction and part (b): Let Sn be the statement that zn = z n.
S1 is true because z1 = z = z 1. Assume Sk is true, that is zk = z k. Then
zk+1 = z1+k = zzk = zzk [part (b) with w = zk] = z1zk = z1+k = zk+1, which shows that Sk+1 is true.
Therefore, by mathematical induction, zn = zn for every positive integer n.
Another proof: Use part (b) with w = z, and mathematical induction.
19. 4x2 + 9 = 0 ⇔ 4x2 = −9 ⇔ x2 = − 94⇔ x = ± − 9
4 = ± 94i = ± 3
2i.
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COMPLEX NUMBERS ■ 11
20. x4 = 1 ⇔ x4 − 1 = 0 ⇔ x2 − 1 x2 + 1 = 0 ⇔ x2 − 1 = 0 or x2 + 1 = 0 ⇔x = ±1 or x = ±i.
21. By the quadratic formula, x2 + 2x + 5 = 0 ⇔
x =−2± 22 − 4(1)(5)
2(1)=−2±√−16
2=−2± 4i2
= −1± 2i.
22. 2x2 − 2x+ 1 = 0 ⇔ x =−(−2)± (−2)2 − 4(2)(1)
2(2)=2±√−4
4=2± 2i4
=1
2± 1
2i
23. By the quadratic formula, z2 + z + 2 = 0 ⇔ z =−1± 12 − 4(1)(2)
2(1)=−1±√−7
2= −1
2±√7
2i.
24. z2 + 12z + 1
4= 0 ⇔ 4z2 + 2z + 1 = 0 ⇔
z =−2± 22 − 4(4)(1)
2(4)=−2±√−12
8=−2± 2√3 i
8= −1
4±√3
4i
25. For z = −3 + 3i, r = (−3)2 + 32 = 3√2 and tan θ = 3−3 = −1 ⇒ θ = 3π
4(since z lies in the second
quadrant). Therefore, −3 + 3i = 3√2 cos 3π4+ i sin 3π
4.
26. For z = 1−√3 i, r = 12 + −√3 2= 2 and tan θ = −√3
1= −√3 ⇒ θ = 5π
3 (since z lies in the fourth
quadrant). Therefore, 1−√3 i = 2 cos 5π3+ i sin 5π
3.
27. For z = 3 + 4i, r =√32 + 42 = 5 and tan θ = 4
3⇒ θ = tan−1 4
3(since z lies in the first quadrant).
Therefore, 3 + 4i = 5 cos tan−1 43+ i sin tan−1 4
3.
28. For z = 8i, r =√02 + 82 = 8 and tan θ = 8
0is undefined, so θ = π
2(since z lies on the positive imaginary axis).
Therefore, 8i = 8 cos π2+ i sin π
2.
29. For z =√3 + i, r =
√3
2+ 12 = 2 and tan θ = 1√
3⇒ θ = π
6⇒ z = 2 cos π
6+ i sin π
6.
For w = 1 +√3 i, r = 2 and tan θ =
√3 ⇒ θ = π
3⇒ w = 2 cos π
3+ i sin π
3. Therefore,
zw = 2 · 2 cos π6+ π
3+ i sin π
6+ π
3= 4 cos π
2+ i sin π
2,
z/w = 22cos π
6 − π3+ i sin π
6 − π3
= cos −π6+ i sin −π
6, and 1 = 1 + 0i = 1(cos 0 + i sin 0) ⇒
1/z = 12cos 0− π
6+ i sin 0− π
6= 1
2cos −π
6+ i sin −π
6. For 1/z, we could also use the formula that
precedes Example 5 to obtain 1/z = 12cos π
6− i sin π
6.
30. For z = 4√3− 4i, r = 4
√3
2+ (−4)2 = √64 = 8 and tan θ = −4
4√3= − 1√
3⇒ θ = 11π
6⇒
z = 8 cos 11π6 + i sin 11π
6. For w = 8i, r =
√02 + 82 = 8 and tan θ = 8
0 is undefined, so θ =π2⇒
w = 8 cos π2+ i sin π
2. Therefore, zw = 8 · 8 cos 11π
6+ π
2+ i sin 11π
6+ π
2= 64 cos π
3+ i sin π
3,
z/w = 88cos 11π
6− π
2+ i sin 11π
6− π
2= cos 4π
3+ i sin 4π
3, and
1 = 1 + 0i = 1(cos 0 + i sin 0) ⇒ 1/z = 18cos 0− 11π
6+ i sin 0− 11π
6= 1
8cos π
6+ i sin π
6.
For 1/z, we could also use the formula that precedes Example 5 to obtain 1/z = 18cos 11π
6− i sin 11π
6.
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12 ■ COMPLEX NUMBERS
31. For z = 2√3 − 2i, r = 2
√3
2+ (−2)2 = 4 and tan θ = −2
2√3= − 1√
3
⇒ θ = −π6⇒ z = 4 cos −π
6+ i sin −π
6. For w = −1 + i, r =
√2,
tan θ = 1−1 = −1 ⇒ θ = 3π
4⇒ w =
√2 cos 3π
4+ i sin 3π
4. Therefore,
zw = 4√2 cos −π
6+ 3π
4+ i sin −π
6+ 3π
4= 4
√2 cos 7π
12+ i sin 7π
12,
z/w = 4√2cos −π
6− 3π
4+ i sin −π
6− 3π
4= 4√
2cos − 11π
12+ i sin − 11π
12
= 2√2 cos 13π
12+ i sin 13π
12, and
1/z = 14 cos −π
6 − i sin −π6
= 14cos π
6+ i sin π
6.
32. For z = 4√3 + i = 4
√3 + 4i, r = 4
√3
2+ 42 =
√64 = 8 and tan θ = 4
4√3= 1√
3⇒ θ = π
6 ⇒
z = 8 cos π6 + i sin π6. For w = −3− 3i, r = (−3)2 + (−3)2 = √18 = 3√2 and
tan θ = −3−3 = 1 ⇒ θ = 5π
4⇒ w = 3
√2 cos 5π
4+ i sin 5π
4. Therefore,
zw = 8 · 3√2 cos π6+ 5π
4+ i sin π
6+ 5π
4= 24
√2 cos 17π
12+ i sin 17π
12,
z/w = 8
3√2cos π
6− 5π
4+ i sin π
6− 5π
4= 4
√2
3cos − 13π
12+ i sin − 13π
12, and
1/z = 18 cos
π6 − i sin π
6 .
33. For z = 1 + i, r =√2 and tan θ = 1
1= 1 ⇒ θ = π
4⇒ z =
√2 cos π
4+ i sin π
4. So by
De Moivre’s Theorem,
(1 + i)20 =√2 cos π
4+ i sin π
4
20
= 21/220
cos 20 ·π4+ i sin 20 ·π
4
= 210(cos 5π + i sin 5π) = 210[−1 + i(0)] = −210 = −1024
34. For z = 1−√3 i, r = 12 + −√3 2= 2 and tan θ = −√3
1= −√3 ⇒ θ = 5π
3⇒
z = 2 cos 5π3 + i sin 5π3 . So by De Moivre’s Theorem,
1−√3 i5
= 2 cos 5π3+ i sin 5π
3
5= 25 cos 5 · 5π
3+ i sin 5 · 5π
3
= 25 cos π3+ i sin π
3= 32 1
2 +√32i = 16 + 16
√3 i
35. For z = 2√3 + 2i, r = 2
√3
2+ 22 =
√16 = 4 and tan θ = 2
2√3= 1√
3⇒ θ = π
6⇒
z = 4 cos π6 + i sin π
6. So by De Moivre’s Theorem,
2√3 + 2i
5
= 4 cos π6+ i sin π
6
5= 45 cos 5π
6+ i sin 5π
6= 1024 −
√32+ 1
2i = −512√3 + 512i
36. For z = 1− i, r =√2 and tan θ = −1
1= −1 ⇒ θ = 7π
4⇒ z =
√2 cos 7π
4+ i sin 7π
4⇒
(1− i)8 =√2 cos 7π
4+ i sin 7π
4
8
= 24 cos 8·7π4+ i sin 8·7π
4
= 16(cos 14π + i sin 14π) = 16(1 + 0i) = 16
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COMPLEX NUMBERS ■ 13
37. 1 = 1 + 0i = 1 (cos 0 + i sin 0). Using Equation 3 with r = 1, n = 8, and θ = 0, we have
wk = 11/8 cos
0 + 2kπ
8+ i sin
0 + 2kπ
8= cos
kπ
4+ i sin
kπ
4, where k = 0, 1, 2, . . . , 7.
w0 = 1(cos 0 + i sin 0) = 1, w1 = 1 cos π4 + i sin π4= 1√
2+ 1√
2i,
w2 = 1 cosπ2+ i sin π
2= i, w3 = 1 cos 3π4 + i sin 3π
4= − 1√
2+ 1√
2i,
w4 = 1(cosπ + i sinπ) = −1, w5 = 1 cos 5π4 + i sin 5π4
= − 1√2− 1√
2i,
w6 = 1 cos3π2+ i sin 3π
2= −i, w7 = 1 cos 7π4 + i sin 7π
4= 1√
2− 1√
2i
38. 32 = 32 + 0i = 32 (cos 0 + i sin 0). Using Equation 3 with r = 32, n = 5, and θ = 0, we have
wk = 321/5 cos
0 + 2kπ
5+ i sin
0 + 2kπ
5= 2 cos 2
5πk + i sin 2
5πk , where k = 0, 1, 2, 3, 4.
w0 = 2(cos 0 + i sin 0) = 2
w1 = 2 cos2π5 + i sin 2π
5
w2 = 2 cos4π5 + i sin 4π
5
w3 = 2 cos6π5+ i sin 6π
5
w4 = 2 cos8π5+ i sin 8π
5
39. i = 0 + i = 1 cos π2+ i sin π
2. Using Equation 3 with r = 1, n = 3, and θ = π
2, we have
wk = 11/3 cos
π2+ 2kπ
3+ i sin
π2+ 2kπ
3, where k = 0, 1, 2.
w0 = cos π6+ i sin π
6=√32+ 1
2i
w1 = cos 5π6+ i sin 5π
6= −
√32+ 1
2i
w2 = cos 9π6+ i sin 9π
6= −i
40. 1 + i =√2 cos π4 + i sin π
4. Using Equation 3 with r =
√2, n = 3, and θ = π
4, we have
wk =√2
1/3cos
π4 + 2kπ
3+ i sin
π4+ 2kπ
3, where k = 0, 1, 2.
w0 = 21/6 cos π
12 + i sin π12
w1 = 21/6 cos 3π
4+ i sin 3π
4= 21/6 − 1√
2+ 1√
2i = −2−1/3 + 2−1/3i
w2 = 21/6 cos 17π
12+ i sin 17π
12
41. Using Euler’s formula (6) with y = π2 , we have e
iπ/2 = cos π2+ i sin π
2= 0 + 1i = i.
42. Using Euler’s formula (6) with y = 2π, we have e2πi = cos 2π + i sin 2π = 1.
43. Using Euler’s formula (6) with y = π
3, we have eiπ/3 = cos π
3+ i sin
π
3=1
2+
√3
2i.
44. Using Euler’s formula (6) with y = −π, we have e−iπ = cos(−π) + i sin(−π) = −1.
45. Using Equation 7 with x = 2 and y = π, we have e2+iπ = e2eiπ = e2(cosπ + i sinπ) = e2(−1 + 0) = −e2.
46. Using Equation 7 with x = π and y = 1, we have eπ+i = eπ · e1i = eπ(cos 1 + i sin 1) = eπ cos 1 + (eπ sin 1)i.
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14 ■ COMPLEX NUMBERS
47. Take r = 1 and n = 3 in De Moivre’s Theorem to get
[1(cos θ + i sin θ)]3 = 13(cos 3θ + i sin 3θ)
(cos θ + i sin θ)3 = cos 3θ + i sin 3θ
cos3 θ + 3 cos2 θ (i sin θ) + 3(cos θ)(i sin θ)2 + (i sin θ)3 = cos 3θ + i sin 3θ
cos3 θ + 3cos2 θ sin θ i− 3 cos θ sin2 θ − sin3 θ i= cos 3θ + i sin 3θ
cos3 θ − 3 sin2 θ cos θ + 3 sin θ cos2 θ − sin3 θ i= cos 3θ + i sin 3θ
Equating real and imaginary parts gives cos 3θ = cos3 θ − 3 sin2 θ cos θ and sin 3θ = 3 sin θ cos2 θ − sin3 θ48. Using Formula 6,
eix + e−ix = (cosx+ i sinx) + [cos(−x) + i sin(−x)]= cosx+ i sinx+ cosx− i sinx = 2cosx
Thus, cosx = eix + e−ix
2.
Similarly,
eix − e−ix = (cosx+ i sinx)− [cos(−x) + i sin(−x)]= cosx+ i sinx− cosx− (−i sinx) = 2i sinx
Therefore, sinx = eix − e−ix
2i.
49. F (x) = erx = e(a+bi)x = eax+bxi = eax(cos bx+ i sin bx) = eax cos bx+ i(eax sin bx) ⇒
F 0(x) = (eax cos bx)0 + i(eax sin bx)0
= (aeax cos bx− beax sin bx) + i(aeax sin bx+ beax cos bx)
= a [eax(cos bx+ i sin bx)] + b [eax(− sin bx+ i cos bx)]
= aerx + b eax i2 sin bx+ i cos bx
= aerx + bi [eax(cos bx+ i sin bx)] = aerx + bierx = (a+ bi)erx = rerx
50. (a) From Exercise 49, F (x) = e(1+i)x ⇒ F 0(x) = (1 + i)e(1+i)x. So
e(1+i)x dx =1
1 + iF 0(x) dx =
1
1 + iF (x) + C =
1− i
2F (x) +C =
1− i
2e(1+i)x +C
(b) e(1+i)x dx = exeix dx = ex(cosx+ i sinx) dx = ex cosxdx+ i ex sinxdx (1). Also,
1− i
2e(1+i)x = 1
2e(1+i)x − 1
2ie(1+i)x = 1
2ex+ix − 1
2iex+ix
= 12ex(cosx+ i sinx)− 1
2iex(cosx+ i sinx)
= 12ex cosx+ 1
2ex sinx+ 1
2iex sinx− 1
2iex cosx
= 12ex(cosx+ sinx) + i 1
2ex(sinx− cosx) (2)
Equating the real and imaginary parts in (1) and (2), we see that ex cosxdx = 12ex(cosx+ sinx) + C and
ex sinxdx = 12ex(sinx− cosx) + C.
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