how to solve $(z-i)^{m}=(z+i)^{m}$ Please give me just a hint not the  whole solution. 

The Problem: Let m be an arbitrary integer. So solve 
  $$(z-i)^{m}=(z+i)^{m},$$ 
  for any $z\in\mathbb{C}.$
Obviously, If $m=1$ there will be no solution and if $m=2$ then we can write, as one case, $z+i=-z+i$ then $z=0$. 

I also tested to write $z=re^{i\theta}$ but no result. 
How should I approach this problem? A small hint will be accepted. 
 A: Note that $z=i$ is not a solution and rewrite
$$\left(\frac{z+i}{z-i}\right)^m=1,$$
then
$$\frac{z+i}{z-i}=\omega^k$$ where $\omega$ is a primitive root of $1$, and $k=0,\cdots m-1$.
From this,
$$z=i\frac{\omega^k+1}{\omega^k-1}=i\frac{(\omega^k+1)(\omega^{-k}-1)}{(\omega^k-1)(\omega^{-k}-1)}=i\frac{\omega^{-k}-\omega^k}{2-\omega^k-\omega^{-k}}=\frac{\Im(\omega^k)}{2(1-\Re(\omega^k))}=\frac{\sin\phi}{2(1-\cos\phi)}$$
where $k\phi=2\pi$.
A: Taking modulo: $|z-i|=|z+i|$ and if $z=a+bi$ then $a^2+(b-1)^2=a^2+(b+1)^2$. This implies that $b=0$, ie $z\in \Bbb{R}$.
Now, because $z\in \Bbb{R}$, then $z+i$ and $z-i$ have argument $\alpha$ and $-\alpha$ respectively and $m\cdot\alpha=m\cdot (-\alpha)+2k\pi$ if and only if $m\alpha=k\pi$ with $k\in\Bbb{Z}$. Can you finish with this?
A: Using the result
$$
1 = e^{\frac{\pi}{2}i+2k\pi i}
$$
We have
$$
\frac{z-i}{z+i} = e^{\frac{\pi}{2m}i+\frac{2k\pi}{m}i}
$$
and then
$$
z = -\frac{i \left(1+e^{\frac{2 i \pi  k}{m}+\frac{i \pi }{2 m}}\right)}{-1+e^{\frac{2 i \pi  k}{m}+\frac{i \pi }{2 m}}} = -\cot \left(\frac{4 \pi  k+\pi }{4 m}\right)
$$
