Proof by induction of Complex Numbers $(Z^n)^*=(Z^*)^n$

Question

Hi I have asked a few questions here before and people have been really helpful, just wondering if someone could help me out with this proof by induction question.

Simply: $(Z^n)^*=(Z^*)^n$ for all positive integers $n$.

Thank you in advance

Answer 1

Let $z=x+yi$.

For $n=2$: $$ (z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*. $$ Assume that it holds for $k=n-2$, i.e., $$ (z^*)^{n-2} = (z^{n-2})^*. $$ Thus, for $(z^*)^n$ $$ (z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*. $$