Difference of two complex polynomials Prove that for any integer $m>1$, $(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}z^2+a^2\cot^2{\frac{k\pi}{2m}}$
I started off by expanding the left hand side using the binomial therem and noticed that some terms cancelled out and I ended up with the expression below for the LHS
$2\sum_{k=1}^{m-1}$$2m\choose{2k-1}$$z^{2m-2k+1}$$a^{2k-1}$
I am stuck as to where to go past this. Some help would be appreciated. Thanks

We follow the comments and consider the polynomial
\begin{align*}
u^{2m}=1\qquad\text{ with roots }\qquad \exp\left(\frac{k\pi i}{m}\right), 1\leq k\leq 2m
\end{align*}
The function
\begin{align*}
u(z)=\frac{z-a}{z+a}
\end{align*}
has the inverse function
\begin{align*}
z(u)=-a\frac{u+1}{u-1}
\end{align*}

We conclude the roots of the equation
  \begin{align*}
\left(\frac{z-a}{z+a}\right)^{2m}=1\tag{1}
\end{align*}
  are if $-m+1\leq k\leq m-1$
  \begin{align*}
-a\cdot\frac{\exp\left(\frac{k\pi i}{m}\right)+1}{\exp\left(\frac{k\pi i}{m}\right)-1}
=-ai\cdot\frac{\frac{1}{2}\left(\exp\left(\frac{k\pi i}{2m}\right)+\exp\left(-\frac{k \pi i}{2m}\right)\right)}{\frac{1}{2i}\left(\exp\left(\frac{k\pi i}{2m}\right)-\exp\left(-\frac{k \pi i}{2m}\right)\right)}
=ia\cot \frac{k\pi}{2m}
\end{align*}
  together with the root $z=0$.

We can now write (1) in polynomial form and as product of factors

\begin{align*}
(z+a)^{2m}-(z-a)^{2m}&=Az\prod_{k=1}^{m-1}\left[\left(z-ia\cot \frac{k\pi}{2m}\right)\left(z+ia\cot \frac{k\pi}{2m}\right)\right]\\
&=Az\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)
\end{align*}
  with $A$ constant.

In order to determine $A$ it is convenient to calculate the coefficient of $z^{2m-1}$, denoted with $[z^{2m-1}]$. We obtain
\begin{align*}
[z^{2m-1}]&\left((z+a)^{2m}-(z-a)^{2m}\right)
=\binom{2m}{2m-1}a-\binom{2m}{2m-1}(-a)=4ma\\
[z^{2m-1}]&Az\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)=A
\end{align*}
We finally obtain with $A=4ma$

\begin{align*}
\color{blue}{(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)}
\end{align*}

