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I have to prove that the following holds. A hint to use complex numbers has been given. I have tried to make a start but not to any result.

$$\prod_{k=1}^{n}\tan\left(\frac{\pi k}{2n+1}\right)=\sqrt{2n+1}$$

My Attempt:

Let us consider $(2n+1)^{\text{th}}$ roots of unity, $z_k=\exp\left(\frac{2k\pi i}{2n+1}\right)$. We can rewrite the product in terms of $\arg(z_k)$ as $\prod_{k=1}^{n}\tan\left(\frac{1}{2}\arg(z_k)\right)$. Or equivalently so, if we consider $(4n+2)^{\text{th}}$ roots of unity, we get this product as $\prod_{k=1}^{n}\tan(\arg(\zeta _k))$, where $\zeta_k= \exp\left(\frac{2k\pi i}{4n+2}\right)$.

I know it can be proved by proving the expression for $\prod\sin(\frac{1}{2}\arg(z_k))$. But I was wondering, is there a way to use telescopic products or purely the geometry of the complex roots of unity to arrive at this

amWhy
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Paras Khosla
  • 6,481

3 Answers3

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We are going to prove that $$\displaystyle \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}. $$ Considering the roots of the following polynomial equation of degree $2m$ $$\displaystyle (z+1)^{2 m+1}-(z-1)^{2 m+1}=0 \tag{$*$} $$ $$\displaystyle \left(\frac{z-1}{z+1}\right)^{2 m+1}=1 $$ Taking $2m+1$ th roots, we have

$$ \frac{z-1}{z+1} =\cos \frac{2 k \pi}{2 m+1}+i \sin \frac{2 k \pi}{2 m+1} =e^{\frac{2 k \pi}{2 m+1} i}, $$ where $k=1,2, \ldots , 2 m$.

Hence the $2m$ complex roots of $(*)$ are

$ \displaystyle z =\frac{e^{\frac{2 k \pi}{2 m+1} i}+1}{1-e^{\frac{2 k \pi}{2 m+1} i}} =\frac{e^{\frac{k \pi}{2 m+1} i}-e^{-\frac{k \pi}{2 m+1} i}}{e^{-\frac{k\pi}{2 m+1} i}-e^{\frac{k \pi}{2 m+1} i}} \tag*{} $ $\displaystyle \qquad =\frac{2 \cos \left(\frac{k \pi}{2 m+1}\right)}{-2 i\sin \left(\frac{k \pi}{2 m+1}\right)} =i \cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $

$\displaystyle \because \text { product of roots of }(*) =\frac{\text{constant in (*)}}{\text{leading coefficient of (*)}}\tag*{} $ $\therefore \displaystyle \prod_{k=1} ^{2 m}\left[i \cot \left(\frac{k \pi}{2 m+1}\right)\right]=\frac{2}{2(2 m+1)} \tag*{} \\$ $\displaystyle i^{2 m} \prod_{k=1}^ {2 m}\cot \left(\frac{k \pi}{2 m+1}\right)=\frac{1}{ 2 m+1} \tag*{} $ $\because \displaystyle \cot \frac{(2 m+1-k)\pi}{2 m+1} =-\cot \left(\frac{k \pi}{2 m+1}\right)\tag*{} $ $\therefore\displaystyle (-1)^{m}(-1) ^m \prod_{k=1}^ { m}\left[ \cot ^{2} \frac{k \pi}{2 m+1}\right]=\frac{1}{2 m+1} \tag*{} $ $\displaystyle \left(\prod_{k=1}^{m} \cot \frac{k \pi}{2 m+1}\right)^{2}=\frac{1}{2 m +1 } \tag*{} \\$ $$\displaystyle \quad \prod_{k=1}^{m}\left(\tan \frac{k \pi}{2 m+1}\right)=\sqrt{2 m+1}$$

Lai
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3

Building on @WETutorialSchool's hint, start from the identity$$i\tan\frac{\theta}{2}=\frac{2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}}{2\cos\frac{\theta}{2}\exp\frac{i\theta}{2}}=\frac{\exp i\theta-1}{\exp i\theta+1}.$$Define $z:=\exp\frac{2i\pi}{2n+1}$ so$$\frac{z^k-1}{z^k+1}=i\tan\frac{k\pi}{2n+1},\,\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}=\frac{z^{-k}-1}{z^{-k}+1}=\frac{1-z^k}{1+z^k}=-i\tan\frac{k\pi}{2n+1}.$$Since $\prod_{k=1}^n\tan\frac{k\pi}{2n+1}$ is a product of the positive tangents of $n$ acute angles,$$\prod_{k=1}^n\tan\frac{k\pi}{2n+1}=\sqrt{\prod_{k=1}^ni\tan\frac{k\pi}{2n+1}\cdot-i\tan\frac{k\pi}{2n+1}}=\sqrt{\prod_{k=1}^n\frac{z^k-1}{z^k+1}\frac{z^{2n+1-k}-1}{z^{2n+1-k}+1}}=\sqrt{\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}}.$$To prove $\prod_{k=1}^{2n}\frac{z^k-1}{z^k+1}=2n+1$, consider the values of $z^k+1,\,0\le k\le 2n$. They are the roots of $(w-1)^{2n+1}-1$, so their product is $-1$ times this polynomial's constant term, i.e. $2$. In other words, $\prod_{k=0}^{2n}(z^k+1)=2$ and $\prod_{k=1}^{2n}(z^k+1)=1$. Similarly, $\prod_{k=1}^{2n}(z^k-1)$ is the product of the roots of $\frac{(w+1)^{2n+1}-1}{w}$, so is equal to its constant term, $\binom{2n+1}{1}$.

J.G.
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2

Let $ n $ be a positive integer.

Consider the Following polynomial : $ P_{n}=\left(X-\mathrm{i}\right)^{2n+1}+\left(X+\mathrm{i}\right)^{2n+1} $, whose zeros can be given by : \begin{aligned} P_{n}\left(z\right)=0\iff \left(z-\mathrm{i}\right)^{2n+1}=\left(-z-\mathrm{i}\right)^{2n+1}&\iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tiny \ \normalsize z-\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\left(-z-\mathrm{i}\right),\ \ \ \ 0\leq k\leq 2n \\ &\iff z\left(1+\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\right)=\mathrm{i}\left(1-\mathrm{e}^{\mathrm{i}\frac{2k\pi}{2n+1}}\right), \ \ \ \ \ 0\leq k\leq 2n\\ &\iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \tiny \ \normalsize z=\tan{\left(\frac{k\pi}{2n+1}\right)}, \ \ \ 0\leq k\leq 2n\end{aligned}

Since $ P_{n} $ can be developed as following :

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{n}=\sum\limits_{k=0}^{2n+1}{\binom{2n+1}{k}z^{2n+1-k}\left(-\mathrm{i}\right)^{k}}+\sum\limits_{k=0}^{2n+1}{\binom{2n+1}{k}z^{2n+1-k}\,\mathrm{i}^{k}}=2\sum\limits_{k=0}^{n}{\left(-1\right)^{k}\binom{2n+1}{2k}z^{2n+1-2k}} $

We have $ P_{n}=2\prod\limits_{k=0}^{2n}{\left(X-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)} $, and thus : $$ \left(\forall z\in\mathbb{C}\right),\ \prod_{k=1}^{2n}{\left(z-\tan{\left(\frac{k\pi}{2n+1}\right)}\right)}=\sum_{k=0}^{n}{\left(-1\right)^{k}\binom{2n+1}{2k}z^{2n-2k}} $$

For $ z=0 $, we get $ \prod\limits_{k=1}^{2n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}=\left(-1\right)^{n}\left(2n+1\right) \cdot $

Since $ \prod\limits_{k=1}^{2n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}=\prod\limits_{k=1}^{n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}\prod\limits_{k=n+1}^{2n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}=\prod\limits_{k=1}^{n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}\prod\limits_{k=1}^{n}{\tan{\left(\frac{\left(2n+1-k\right)\pi}{2n+1}\right)}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(-1\right)^{n}\left(\prod\limits_{k=1}^{n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}\right)^{2} $

We have : $$ \prod_{k=1}^{n}{\tan{\left(\frac{k\pi}{2n+1}\right)}}=\sqrt{2n+1} $$

CHAMSI
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