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

Lightning Apr 14, 2019

#1**+1 **

(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

CPhill Apr 14, 2019