Originally Posted by

**Blizzardy** Hi guys, need help with a question:

Given that z = 2 + 2i, express z* in the exponential form. Hence find the set of the positive interger n for which (z*)^n is purely imaginary.

z* = 2^(3/2) e^[i(-pi/4)]

(z*)^n = 2^(3n/2) e^[i(-npi/4)]

(i) Given that (z*)^n is purely imaginary,

-npi/4 = pi/2, 3pi/2, 5pi/2, ...

-npi/4 = (2m + 1)(pi/2) where m 'is an element of all intergers, and m > than or = 0'

This is incorrect. It should be, $\displaystyle -\frac{n\pi}{4}=2k\pi\pm\frac{\pi}{2}\mbox{ where }k\in Z$. I have shown how to obtain this below.

n = -(4m+2)

(But my ans was wrong)

Anyway, if the question wants to find the set of intergers for n for which (z*)^n is real, can I get the answer simply by letting '-npi/4 = 0, pi, 2pi, 3pi,...' ?

Thanks in advance!