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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.

Jean
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2 Answers2

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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!

Community
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Mark Viola
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  • nope, this is brilliant, thank you very much! – Jean Jul 22 '15 at 20:32
  • You're welcome. And my pleasure. And thank you for the nice compliment. I'm recovering at home from surgery 3 weeks ago. Your kind words just made my day! – Mark Viola Jul 22 '15 at 20:41
  • Hope you get well soon! :) – Jean Jul 22 '15 at 20:42
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    As an alternative to binomial expansion, you can obtain $C$ by dividing both sides of equation (3) by $2a$, then taking $a\to 0$ and recognizing the LHS as a derivative. – Semiclassical Jul 22 '15 at 20:42
  • @Semiclassical YES, I know noticed that too! – Jean Jul 22 '15 at 20:44
  • @semiclassical Yes you're absolutely correct. I thought using the binomial expasion was quick and aligned with the OP's work shown in the post. – Mark Viola Jul 22 '15 at 20:46
  • here the power is 2m. So it should have 2m roots. $$\pm1,\pm2,.....,\pm(m-1)$$ that is 2m-2 amount of roots also z=0 is a root, a total of 2m-1 roots. Where is the last one? Is it because $Z^{2}$ is present in the factorization? – ALvi1995 Jun 27 '21 at 18:08
  • @ALvi1993 Actually, the polynomial $(z+a)^{2n}-(z-a)^{2m}$ is of order $2m-1$, not $2m$. – Mark Viola Jun 27 '21 at 18:46
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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.

heropup
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    No this was really nice, I learned the $x^{n}-y^{n}$ identity, so it was very useful, thank you! – Jean Jul 22 '15 at 19:31