Determine all n element N, so that (1-i)^n is natural

I've done following steps thus far:

define z:= 1-i; then z=sqrt(2) * e^((-pi/4)i)
(1-i)^n = z^n = (sqrt(2)*e^((-pi/4)i)^n = sqrt(2)^n * e^((-pi/4)in
That gives two conditions:
1) sqrt(2)^n needs to be element N -> n=2k, k element N
2) e^((-pi/4)in needs to be element N -> pi/4 * n = pi * k -> n=4k, k element N


At this point I'm lost, though. I know the solution from straight thinking - n needs to be 8*k, but how do I get to that point from where I am with my steps so far? E.g. From the two conditions I have here, n could also be 4, but (1-i)^n is -4 which is not element N.

Cheers Anton

• So your two conditions are: (1) $n$ must be a multiple of $2$, and (2) $n$ must be a multiple of $4$. Notice that (2) implies (1).
– user169852
Commented Dec 21, 2020 at 10:26
• $n$ must be a multiple of $2$ and a multiple of $4$. Why do you say that these two conditions contradict each other? Commented Dec 21, 2020 at 10:32
• Right, I expressed myself wrongly. But how do I write it down? Both conditions - n a multiple of 2 and n a multiple of 4 - are also true for n=4 - but then (1-i)^n is -4 which is not element N... Commented Dec 21, 2020 at 10:52
• $(1-i)^4=-4$ therefore $(1-i)^8=16$ Commented Dec 21, 2020 at 11:06
• Welcome to Maths SE. Please use MathJax. Commented Dec 21, 2020 at 12:20

2) e^((-pi/4)in needs to be element N -> pi/4 * n = 2*pi * k -> n=8*k, k element N