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*}
