+1245 Suppose that a and b are positive integers such that (a-bi)^2 = 8-6i. What is a-bi?
+129852 (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
