Solve in C :
n is in N and is the conjucate of .
$\displaystyle (z-1)^n=(\overline{z}+1)^n\Rightarrow\left(\frac{z-1}{\overline{z}+1}\right)^n=1$
Apply the modulus to both members. Then
$\displaystyle \left|\frac{z-1}{\overline{z}+1}\right|=1\Rightarrow|z-1|=|\overline{z}+1|$
Let $\displaystyle z=x+yi$
$\displaystyle (x-1)^2+y^2=(x+1)^2+y^2\Rightarrow x=0, \ y\in\mathbb{R}$
So, $\displaystyle z=yi$
Replace y in the equation: $\displaystyle (-1+yi)^n=(1-yi)^n\Rightarrow(-1+yi)^n=(-1)^n(-1+yi)^n$
If n is even the equation is verified for every real y.
If n is odd the equation has no solution.