Complex number question...?

Question

Show that
(1/2(a+(a^2+b^2)^-2))^-2+i(1/2(-a+(a^2+b^2)^-2))^-2
squares to a+ib.

Correction:
(1/2(a+(a^2+b^2)^-2))^1/2+
i(1/2(-a+(a^2+b^2)^-2))^1/2

Answer 1

You still have it wrong. It should be
(½[a + √(a² + b²)])^½ + i(½[-a + √(a² + b²)])^½
squares to a + bi

Let c = (½[a + √(a² + b²)])^½, d = (½[a + √(a² + b²)])^½

Then you want to show that (c + di)² = a + bi

(c + di)² = c² + 2cdi + (di)² = c² - d² + 2cdi

c² = ½[a + √(a² + b²)]
d² = ½[-a + √(a² + b²)]
c² - d² =½[a + √(a² + b²)] - ½[-a + √(a² + b²)] = a

2cd = 2(½[a + √(a² + b²)])^½ (½[-a + √(a² + b²)])^½
= 2{(½[a + √(a² + b²)])(½[-a + √(a² + b²)])}^½
= 2{(¼)[a + √(a² + b²)][-a + √(a² + b²)])}^½
= 2{(¼)(-a² + a√(a² + b²) - a√(a² + b²) + (a² + b²))}^½
= 2{(¼)(-a² + (a² + b²))}^½
= 2√(¼ b²)
= b