Prove…

Prove: $\ (z+a)^{2m}-(z-a)^{2m}=4maz\displaystyle\prod_{k=1}^{m-1}\left[z^2+a^2\cot^2\left(\dfrac{k\pi}{2m}\right)\right]$.

Step 1 · The degree of $f$ is $2m-1$, not $2m$

Expanding $(z\pm a)^{2m}$, the $z^{2m}$ terms cancel between the two expansions, leaving as leading term:

$$ f(z)=(z+a)^{2m}-(z-a)^{2m} = 2\binom{2m}{1}az^{2m-1}+\cdots = 4maz^{2m-1}+\cdots $$

so $f$ is a polynomial of degree $2m-1$.

Step 2 · Find the roots

$f(z)=0$ (with $z\neq a$) means $\left(\dfrac{z+a}{z-a}\right)^{2m}=1$, so $\dfrac{z+a}{z-a}=\omega$ for some $2m$-th root of unity $\omega=e^{i\pi k/m}$, $k=0,1,\ldots,2m-1$. Solving for $z$:

$$ z+a=\omega(z-a) \quad\Longrightarrow\quad z(1-\omega)=-a(1+\omega) \quad\Longrightarrow\quad z=\frac{a(1+\omega)}{\omega-1} $$

Step 3 · Simplify using half-angle exponentials

Writing $\omega=e^{i\pi k/m}$: $\ 1+\omega=2\cos\!\left(\tfrac{k\pi}{2m}\right)e^{ik\pi/(2m)}$ and $\omega-1=2i\sin\!\left(\tfrac{k\pi}{2m}\right)e^{ik\pi/(2m)}$, so:

$$ z_k = \frac{a\cdot2\cos\frac{k\pi}{2m}}{2i\sin\frac{k\pi}{2m}} = -ia\cot\left(\frac{k\pi}{2m}\right) $$

Step 4 · Which $k$ give valid finite roots

$k=0$ gives $\omega=1$, which forces $2a=0$ — not a valid finite root for $a\neq0$ (this corresponds to the "missing" top-degree term, consistent with $\deg f=2m-1$ rather than $2m$). For $k=1,\ldots,2m-1$, each $z_k=-ia\cot\frac{k\pi}{2m}$ is a genuine root, giving $2m-1$ roots in total — matching the degree exactly.

Step 5 · Pair the roots symmetrically

Since $\cot\!\left(\pi-\theta\right)=-\cot\theta$, for $k=1,\ldots,m-1$:

$$ z_{2m-k}=-ia\cot\left(\pi-\frac{k\pi}{2m}\right)=ia\cot\left(\frac{k\pi}{2m}\right)=-z_k $$

So the roots come in $m-1$ pairs $\pm ia\cot\frac{k\pi}{2m}$ ($k=1,\ldots,m-1$), plus the single root $z_m=-ia\cot\frac\pi2=0$ — a total of $2(m-1)+1=2m-1$ roots, confirming Step 4.

Step 6 · Assemble the factorization

Grouping each pair $\pm z_k$ into $z^2-z_k^2$, and noting $z_k^2=(-ia\cot\frac{k\pi}{2m})^2=-a^2\cot^2\frac{k\pi}{2m}$, so $z^2-z_k^2=z^2+a^2\cot^2\frac{k\pi}{2m}$:

$$ f(z)=Cz\prod_{k=1}^{m-1}\left[z^2+a^2\cot^2\left(\frac{k\pi}{2m}\right)\right] $$

Step 7 · Match the leading coefficient

Both sides have leading term $Cz^{2m-1}$; comparing with Step 1 ($4maz^{2m-1}$) gives $C=4ma$:

$$ \boxed{(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}\left[z^2+a^2\cot^2\left(\frac{k\pi}{2m}\right)\right]} $$

Verification

Checked symbolically (via expansion) for $m=2,3,4$: both sides agree identically as polynomials in $z$ with parameter $a$.