Prove that $f=(x+i)^{10}+(x-i)^{10}$ have all real roots 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:


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*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 $\mathbb{R}$


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



 A: $(x+i)^{10} = -(x-i)^{10} = i^{10}(x-i)^{10} = (i(x-i))^{10} \to \left(\dfrac{x+i}{ix+1}\right)^{10} = 1$. Can you continue??
A: $(x+i)^{10}+(x-i)^{10}=0$ immediately implies $|x+i|=|x-i|$, which is equivalent to $x \in \mathbb R$ (the real axis is the perpendicular bisector of the line from $i$ to $-i$).
A: Let $a+ib$ be an imaginary root of $f$ , where , $a,b\in \mathbb R$.
Then $$f(a+ib)=0$$
$$\implies (a+i(b+1))^{10}+(a+i(b-1))^{10}=0$$
Expand two binomial series , and separate real part and imaginary parts. Put them separately $0$. Then you obtain the value of $b$, which would be zero, if the problem is proper.
A: 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 $b=0$.
A: You can transform the equation in
$$\left(\frac{x+i}{x-i}\right)^{10}=-1=z^{10}.$$
Then, solving for $x$ and using $|z|=1$,
$$x=i\frac{z+1}{z-1}=i\frac{(z+1)(\overline z-1)}{(z-1)(\overline z-1)}=i\frac{|z|^2-1+\overline z-z}{|z|^2+1-z-\overline z}=\frac{2\Im z}{2-2\Re z}.$$
A: $x=a+bi$ 
$$(x+i)^{10}+(x-i)^{10}=0 \Rightarrow (x+i)^{10}=-(x-i)^{10} \Rightarrow (\frac{x+i}{x-i})^{10}=-1 \\ \Rightarrow \left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=i \\ \Rightarrow \frac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}=i \Rightarrow \frac{a^2-a(b-1)i+a(b+1)i+(b^2-1)}{a^2+(b-1)^2}=i \Rightarrow \frac{(a^2+(b^2-1))+2ai}{a^2+(b-1)^2}=i$$ 
That means that $$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \text{ and } \frac{2a}{a^2+(b-1)^2}=1$$ 
$$\frac{a^2+(b^2-1)}{a^2+(b-1)^2}=0 \Rightarrow a^2+b^2-1=0 \Rightarrow b^2=1-a^2$$ 
$$\frac{2a}{a^2+(b-1)^2}=1 \Rightarrow 2a=a^2+(b-1)^2 \Rightarrow 2a=a^2+b^2-2b+1 $$ 
Solving for $a$ and $b$ we get the following solutions: 
$a=0, b=1$ and $a=1, b=0$ 
We reject $(a, b)=(0, 1)$ because then the denominator would be $0$. 
So, the solution is $(a, b)=(1, 0)$. 
That means that $x=1 \in \mathbb{R}$. 
$$$$ 
Do the same for the case $\left (\frac{a+(b+1)i}{a+(b-1)i}\right )^{10}=i^{10} \Rightarrow \frac{a+(b+1)i}{a+(b-1)i}=-i$
