Let $n\in\mathbb{Z}$. What are all solutions, for $z\in\mathbb{C}$, of $ z^n=\overline{z} $?
To solve it, I tryed to write the term in Polar-Form and than take the logarithm. Because of the "ln(r)" term I was unable to find a solution.
I also tryed to write $ z^n $ as a Binomial, but this was not helpful at all.
This question was asked as an early exercise in Advanced Mathematics 2.
On
In short, you're asking to solve: $$r^ne^{ni\theta}=re^{-i\theta}$$ That is: $$r^{n-1}e^{(n+1)i\theta}=1+0i$$ $$\rightarrow r^{n-1}(\cos[(n+1)\theta]+i\sin[(n+1)\theta])=1$$ Thus $$r^{n-1}(1)=1$$, since when $\sin (x)=0, \cos(x)=\pm1$, and it can't be $-1$ here.
$\sin[(n+1)\theta]=0$ when $\cos[(n+1)\theta]=1$ $\rightarrow$ $(n+1)\theta=2k\pi$ and $r$ can only be $1$
If any part of this doesn't make full sense, please say so and I'll clear up the confusion.
Hint: Assuming $z=re^{it}\neq 0$ (which may or may not be a solution depending on the sign of $n$), your equation is equivalent to$$r^ne^{int}=re^{-it}$$ $$r^{n-1}e^{i(n+1)t} = 1\tag{uses $r\neq 0$}$$ $$\implies r=1,(n+1)t=2k\pi$$ This is enough for you to enumerate the solutions. You must consider the cases $n=-1$ and $n\neq -1$ separately.