Suppose that a and b are positive integers such that (a+bi)^2 = 3+4i. What is a+bi?
+128656 (a+bi)^2 = 3+4i
(a + bi)^2 = a^2 + 2abi - b^2
Equating coefficients, we have
a^2 - b^2 = 3 (1)
2ab = 4 ⇒ ab = 2 ⇒ b = 2/a (2)
Sub (2) into (1)
a^2 - (2/a)^2 = 4
a^4 - 4 = 3a^2
a^4 - 3a^2 - 4 = 0 factor
(a^2 - 4) (a^2 + 1) = 0
A real solultion will come from
(a^2 - 4) = 0
(a + 2) (a - 2) = 0
Since a is positve, then a = 2
And b = 2/a = 2/2 = 1
a + bi = 2 + i
