TITLE: Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots
QUESTION [8 upvotes]: We have $f=(x+i)^{10}+(x-i)^{10}$ and we need to prove that $f$ have all the roots in $\mathbb{R}$.
Here is all my steps:

Suppose that $z\in\mathbb{R}$ is a root of $f\Rightarrow (z+i)^{10}+(z-i)^{10}=0$ 

Therefore: $f(z)=\sum_{k=0}^{10}\left[\left(\dbinom{10}{k}\cdot z^{10-k}\cdot i^k\right)\left(1+(-1)^k\right)\right]=0$

I don't have ideea how can I prove that $f$ have all real roots

Suppose that $z=a+bi$ and we need to prove that $b=0$:
$$\Rightarrow f(z)=\sum_{k=0}^{10}\left(\dbinom{10}{k}\cdot a^{10-k}\cdot i^k\right) \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow \left[(b+1)^k+(b-1)^k \right]=0$$
$$\Rightarrow b+1=-b+1$$
$$\Rightarrow b=0$$
Therefore $f$ have all the roots in $\mathbb{R}$

Here is a photo with the proof of real axis, I don't know if is correct:

REPLY [2 votes]: Here is another way of showing that all solution are real valued.
$(x+i)^{10}+(x-i)^{10}=0\implies\,\,\left(\frac{x+i}{x-i}\right)^{10}=e^{in\pi}\implies \frac{x+i}{x-i}=e^{in\pi/10} \implies x=\cot(n\pi/20)$
which are real-valued for all integer $n$.

To answer the question regarding the substitution $z=a+ib$ and subsequent analysis, if $(b+1)^k+(b-1)^k=0$, then this implies that $b=-i\cot (n\pi/2k)$ for all integer $n$ and $k=0, 1, 2, \cdots 10$.  
But, since $b$ is presumed to be real valued, then this implies $b=0$ (i.e., only solutions with $n/k$ odd are valid).
However, this assumption (i.e., $(b+1)^k+(b-1)^k=0$) is not the basis for a proof that $b=0$.