Prove $Re(x_i)=-\frac{m}{2}$, where $x_i$ are roots of equation $(x+m)^n-x^n $

Question

I was wondering how to factorize this polynomial $(a+b)^n-a^n$

Suddenly, I found an interesting fact that for this equation $$(x+m)^n-x^n=0 \quad n\in \Bbb N^+$$

The set $\{x_i\}$ denotes the zeros of this equation which has $n-1$ element.

$Re(x)$ denotes the real part of complex number $x$.

Then $R(x_i)=-\frac{m}{2}$

Additionally, if $n$ is not an integer, then there is no solution.

I tried several cases but I don't know how to prove it even using the induction.

Answer 1

Here is a general proof:

Rearrange the equation,

$$\left( 1+ \frac mx\right)^n=1$$

to get the solutions

$$1+\frac {m}{x_k}= e^{i\frac{2\pi k}{n}}$$

with $k=1,2,...n-1$. Explicitly,

$$x_k = \frac{m}{e^{i\frac{2\pi k}{n}}-1}=\frac m2 \frac{e^{-i\frac{\pi k}{n}}}{i\sin\frac{\pi k}{n}}=\frac m2 \left( -1-i \cot \frac{\pi k}{n}\right)$$

Thus,

$$Re(x_k)=-\frac{m}{2}$$

Answer 2

If $(x+m)^n-x^n=0$, then $(x+m)^n=x^n$ and therefore $\lvert x+m\rvert^n=\lvert x\rvert^n$. So, $\lvert x+m\rvert=\lvert x\rvert$. In other words, the distance from $x$ to $0$ is equal to the distance from $x$ to $-m$. And this means that $\operatorname{Re}(x)=-\frac m2$.