TITLE: The solutions of $(z-i)^n+(z+i)^n=z^n$ are real.
QUESTION [14 upvotes]: I have to prove that the polynomial
$$(z-i)^n+(z+i)^n=z^n $$
has only real solutions. 
The equation is equivalent to
$$ 2(z^n-C_n^2z^{n-2}+C_n^4z^{n-4}-...)=z^n .$$

REPLY [2 votes]: Consider the function $p_n:\Bbb{R}\to\Bbb{C}$ given by
$$
p_n(x)=(1+ix)^n.
$$
The argument
$$
\theta_n(x):=\arg p_n(x)=n\arctan x
$$
then increases from $-n\pi/2$ to $+n\pi/2$ as $x$ increases from $-\infty$ to
$+\infty$.
Observation #1: In any interval, where $\theta_n(x)$ either

increases from $-\pi/2+2k\pi$ to $-\pi/3+2k\pi$ for some integer $k$, or
increases from $2k\pi+\pi/3$ to $2k\pi+\pi/2$,

there exists an $x$ such that the real part of $p_n(x)=1/2$.
Proof: The absolute value $p_n(x)$ is always $\ge1$. So if $[a,b]$ is an interval of one of the above types, then at the end corresponding to an odd multiple of $\pi/2$, the real part of $p_n(x)=0$. At the other end the cosine of the argument is $1/2$ and therefore the real part of $p_n(x)\ge1/2$. The claim follows from continuity.
Observation #2: If $p_n(x)$ has real part $=1/2$, then $z=1/x$ is a solution
of the original equation.
Proof: If $p_n(x)=1/2+it$, then
$$
(1+ix)^n+(1-ix)^n=p_n(x)+\overline{p_n}(x)=(\frac12+it)+(\frac12-it)=1.
$$
Dividing this by $x^n$ gives
$$
(1/x+i)^n+(1/x-i)^n=(1/x)^n
$$
proving the claim.
The claim follows from these observations, the known codomain of $\theta_n(x)=n\arctan x$, and the fact that $z=0$ is a solution for all odd $n$. If $n=2m$ is even, then there are exactly $2m$ intervals of the type covered in the first observation. OTOH, if $n=2m+1$ then those same $2m$ intervals are included in the range of $\theta_n(x)$, and we have the extra zero from $z=0$.
In either case we have found $n$ real zeros, so there cannot be others.