For that question, here is the answer the prof wrote.
Find all the solutions of Z^6 = 1.
|z|^6(cos6θ + isin6θ)
|z|^6 = 1
|z| = 1
Therefore:
1(1 +0i) <-- don't get how the prof got to here.
cos6θ = 1 sin6θ = 0
6θ = 0+n(2π)
= n(π/3)
so 0, π/3, 2π/3, π...
Hi guys, just a simple question.
I need to find all solutions of z^6 = 1.
I need to change that to |z|^6(cos 6θ + isin6θ) = 1 and the prof wrote → 1(1+0i). How did the prof do this? The rest of the question I understand. It's just that one step still stumps me. Thanks