Prove that for any integer $m>1$, $\ \ (z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}[z^2+a^2\cot^2(k\pi/2m)]$. 
Prove that for any integer $m>1$, 
  $$(z+a)^{2m}-(z-a)^{2m}=4maz\prod\limits_{k=1}^{m-1}[z^2+a^2\cot^2(k\pi/2m)].$$

This how tried to do it: 


*

*Expand the two brackets on the right hand side and end up with, after cancelling the $a^{2n}$ terms (where the variables are multiples of squares):
$$2\sum\limits_{r=1}^{m} \binom{2m}{2r-1}z^{2m-r}a^{2r-1}=4maz \sum\limits_{r=1}^{m} \frac{1}{2m}\binom{2m}{2r-1}z^{2(m-r)}a^{2(r-1)}$$


But now what? How is that series on the left hand side related to the product in the question? Furthermore this question comes from Schaum's Outline Complex Variables, but I tried and failed at integrating complex numbers into this, so complex numbers method would be appreciated here.
BTW you don't have to provide a full proof, just a few hints should be enough.
 A: Using my answer here we can see that 
$$z=ia\cot(k\pi/2m)\,\,\text{for}\,\,k=\pm 1,\pm 2, \cdot ,\pm m-1 \tag 1$$
comprise $2m-2$ roots of the equation
$$\left(\frac{z-a}{z+a}\right)^{2m}=1 \tag 1$$
We also have the obvious root $z=0$.  Thus, we can rearrange $(2)$ in polynomial form, and write the polynomial in factored form in terms of these roots in $(1)$ as
$$\begin{align}
(z+a)^{2m}-(z-a)^{2m}&=Cz\prod_{k=1}^{m-1}\left(z-ia\cot(k\pi/2m)\right)\left(z+ia\cot(k\pi/2m)\right)\\\\
&=Cz\prod_{k=1}^{m-1}\left(z^2+a^2\cot(k\pi/2m)\right)\tag 3
\end{align}$$
where $C$ is the coefficient on the leading term $z^{2m-1}$.  Note that in $(3)$, we exploited the fact that $\cot x=-\cot (-x)$.
Finally, using the binomial expansion, we find that $C$ is 
$$C=2\binom{2m}{2m-1}a=4ma$$
and therefore, we have 
$$\bbox[5px,border:2px solid #C0A000]{(z+a)^{2m}-(z-a)^{2m}=(4ma)\,z\,\prod_{k=1}^{m-1}\left(z^2+a^2\cot(k\pi/2m)\right)}$$
as was to be shown!
A: No; if you notice that the RHS is a finite product of quadratic factors, then this immediately suggests a factorization in complex conjugate pairs.  The idea is to observe the identity $$x^{2m} - y^{2m} = \prod_{k=1}^{2m} (x - \zeta_{2m}^k y)$$ where $\zeta_{2m} = e^{i \pi/m}$ is a primitive $2m^{\rm th}$ root of unity.  This identity may look more familiar when $y = 1$ and for a general positive integer $2m \to n$; i.e., $$x^n - 1 = \prod_{k=1}^n (x - \zeta_n^k).$$  But since we restrict the power to a positive even integer $2m$, we can pair up linear factors in complex conjugate pairs, and the real-valued factors also pair up naturally:  $$\begin{align*} x^{2m} - y^{2m} &= (x - y)(x + y) \prod_{k=1}^{m-1} (x - \zeta_{2m}^k y)(x - \zeta_{2m}^{-k} y) \\ &= (x^2 - y^2) \prod_{k=1}^{m-1} \bigl( x^2 + y^2 - xy (\zeta_{2m}^k + \zeta_{2m}^{-k}) \bigr) \\ &= (x^2-y^2) \prod_{k=1}^{m-1} \Bigl( x^2 + y^2 - 2xy \cos \frac{k \pi}{m} \Bigr). \end{align*}$$  Now we substitute $x = z + a$, $y = z - a$:  we get $$x^2 - y^2 = 4az,$$ and $$\begin{align*} x^2 + y^2 - 2xy \cos \frac{k \pi}{m} &= 2z^2 + 2a^2 - 2(z^2 - a^2) \cos \frac{k \pi}{m} \\ &=  2 \biggl(1 + \cos \frac{k\pi}{m} \biggr) \Biggl( z^2 + a^2 \biggl(\frac{1 - \cos \frac{k\pi}{m}}{1 + \cos \frac{k\pi}{m}} \biggr)\Biggr) \\ &= 4 \cos^2 \frac{k\pi}{2m} \Bigl( z^2 + a^2 \cot^2 \frac{k \pi}{2m} \Bigr). \end{align*}$$  So all that remains is to show that $$\prod_{k=1}^{m-1} 4 \cos^2 \frac{k \pi}{2m} = m.$$  I leave this as a relatively simple exercise for you to complete.  Hint:  observe $\cos \frac{-k\pi}{2m} = \cos \frac{k\pi}{2m}$.

There is very likely a more elegant and simple approach but this is the solution I came up with off the top of my head.
