complex numbers 1.5 true or false: all numbers are complex numbers
TRANSCRIPT
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Complex numbers 1.5
True or false:
All numbers are Complex numbers
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POD
Consider our final equation from the previous lesson. We determined that x = -2 and x = 1 were solutions.
How many solutions will there be, counting multiplicities and complex solutions?
Are -2 and 1 the only solutions, or are there other solutions as well?
How could we find out?
087 36 xx
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POD
Remember that substitution? Let’s factor with it.
Ooh, a difference of cubes and a sum of cubes. How do you factor those?
0)1)(8(
0)1)(8(
087
087
33
2
36
xx
mm
mm
xx
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PODLet’s do a complete factorization. Finding all the
factors helps us find all the solutions.
We can see our two solutions easily. Do they have a multiplicity greater than one? (CAS does this easily.)
How do we find the other solutions?
0)1)(1)(42)(2(
0)1)(8(
087
22
33
36
xxxxxx
xx
xx
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PODTurns out we have some imaginary solutions.
Use the quadratic formula to find them.
2
31
2
31
2
)1)(1(411
1
312
322
2
122
2
)4)(1(442
42
0)1)(1)(42)(2(
2
2
22
ix
xx
ii
x
xx
xxxxxx
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PODAll six solutions:
2
31
31
1
2
0)1)(1)(42)(2( 22
ix
ix
x
x
xxxxxx
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A brief review of i
What is i?
What would i2 equal?
What about i3 or i8?
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A brief review of i
What is i?
What would i2 equal? -1
What about i3 or i8?
1
1
18
3
i
ii
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A brief review of their form
a + bi(real component) (imaginary
component)
What do you have when a = 0?
What about when b = 0?
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Equivalent complex numbers
a + bi = c + di only if a = c and b = d
Solve for x and y:(2x-4) + 9i = 8 + 3yi
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Equivalent complex numbers
a + bi = c + di only if a = c and b = d
Solve for x and y:(2x-4) + 9i = 8 + 3yi2x-4 = 8 9 = 3yx = 6 y = 3
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Adding complex numbers
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i)
Subtract: (3 + 4i) - (2 - 5i)
Try this on calculators.
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Adding complex numbers
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i) = 5 - i
Subtract: (3 + 4i) - (2 - 5i) = 1 + 9i
Try this on calculators.
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This is how it looks on the TI-84.
(a + bi) + (c + di) = (a + c) + (b + d)i
Add: (3 + 4i) + (2 - 5i)
Subtract: (3 + 4i) - (2 - 5i)
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Multiplying complex numbers
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Multiply: (3 + 4i)(2+5i)
Multiply: (3 - 4i)(2+5i)
Try this on calculators.
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Multiplying complex numbers
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Multiply: (3 + 4i)(2+5i) = -14 + 23i
Multiply: (3 - 4i)(2+5i) = 26 + 7i
Try this on calculators.
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Complex conjugates
What is the complex conjugate of a + bi?
What is the product of a + bi and its complex conjugate?
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Complex conjugates
What is the complex conjugate of a + bi?
What is the product of a + bi and its complex conjugate?
That means the factorization for is
22 ba
22 ba
))(( biabia
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Operations with complex numbers
Express in a + bi form:4(2 + 5i) - (3 - 4i)
(4 - 3i)(2 + i) (3 - 2i)2
i(3 + 2i)2
i51
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Operations with complex numbers
Express in a + bi form:4(2 + 5i) - (3 - 4i) = 5 +24i
(4 - 3i)(2 + i) = 11 – 2i (3 - 2i)2 = 5 - 12i i(3 + 2i)2 = -12 + 5i i51 = i3 = -i
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Rational expressions with complex numbers
Simplify
Hint: What is the complex conjugate of the denominator?
And try this on calculators. How do you get rational coefficients?
7 i3 5i
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Rational expressions with complex numbers
Simplify
i
iii
iii
i
i
i
i
17
16
17
1317
1613
34
3226259
533521
)53(
)53(
)53(
)7(
2
2
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Complex numbers as radical expressions
Multiply
Hint: Rewrite using i.
5 9 1 4
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Complex numbers as radical expressions
Multiply.
Without using i, we’d have three different radicals, and wind up with a different real number component.
i
iii
ii
131
63105
2135
4195
2