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Suppose that a and b are positive integers such that (a-bi)^2 = 8-6i. What is a-bi?

 Apr 14, 2019
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(a - bi)^2 =

 

a^2 - 2abi - b^2    =   8 - 6i

 

(a^2 - b^2) = 8     (1)

 

(-2ab)i  = - 6i

2ab i = 6i

ab  = 3      ⇒   b = 3/a      (2)

 

Sub (2) into (1)

 

a^2 - (3/a)^2  = 8

a^2 - 9/a^2  = 8

a^4 - 9  = 8a^2

a^4 - 8a^2 - 9  = 0

(a^2 - 9) (a^2 + 1)   = 0

 

The second factor produces a non-real for a

a^2 - 9  ⇒  a  = 3

And b = 3/a = 3/3  = 1

 

So  (a, b)  = (3, 1)

 

And

 

a - bi  =

 

3  - i

 

 

cool cool cool

 Apr 14, 2019

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