show that $\prod_{k=1}^{n-1}\left(2\cot{\frac{\pi}{n}}-\cot{\frac{k\pi}{n}}+i\right)$ is purely imaginary number

Question

Show that $$\prod_{k=1}^{n-1}\left(2\cot{\dfrac{\pi}{n}}-\cot{\dfrac{k\pi}{n}}+i\right)$$ is purely imaginary number

where $i^2=-1$

where $n=2$ it is clear $$2\cot{\dfrac{\pi}{n}}-\cot{\dfrac{k\pi}{n}}+i=2\cot{\dfrac{\pi}{2}}-\cot{\dfrac{\pi}{2}}+i=i$$ is purely imaginary number

Answer 1

First we have

$$\bigg(2\cot\bigg(\frac{\pi}{n}\bigg)-\cot\bigg(\frac{k\pi}{n}\bigg)+i\bigg)\sin\bigg(\frac{\pi}{n}\bigg)\sin\bigg(\frac{k\pi}{n}\bigg)=\sin\bigg(\frac{(k-1)\pi}{n}\bigg)+\sin\bigg(\frac{k\pi}{n}\bigg)\exp\bigg(i\frac{\pi}{n}\bigg)$$

So it suffices to show that the following product $P$ is purely imaginary:

$$P=\prod_{k=1}^{n-1}\bigg(\sin\bigg(\frac{(k-1)\pi}{n}\bigg)+\sin\bigg(\frac{k\pi}{n}\bigg)\exp\bigg(i\frac{\pi}{n}\bigg)\bigg)$$

We have

$$\overline{P}=\prod_{k=1}^{n-1}\bigg(\sin\bigg(\frac{(k-1)\pi}{n}\bigg)+\sin\bigg(\frac{k\pi}{n}\bigg)\exp\bigg(i\frac{-\pi}{n}\bigg)\bigg)$$

Hence

$$\overline{P}=\exp\bigg(-i\frac{\pi}{n}(n-1)\bigg)\prod_{k=1}^{n-1}\bigg(\sin\bigg(\frac{(k-1)\pi}{n}\bigg)\exp\bigg(i\frac{\pi}{n}\bigg)+\sin\bigg(\frac{k\pi}{n}\bigg)\bigg)$$

Hence, writing $m=n-k$

$$\overline{P}=-\exp\bigg(i\frac{\pi}{n}\bigg)\prod_{m=1}^{n-1}\bigg(\sin\bigg(\frac{(n-m-1)\pi}{n}\bigg)\exp\bigg(i\frac{\pi}{n}\bigg)+\sin\bigg(\frac{(n-m)\pi}{n}\bigg)\bigg)$$

and

$$\overline{P}=-\exp\bigg(i\frac{\pi}{n}\bigg)\prod_{m=1}^{n-1}\bigg(\sin\bigg(\frac{(m+1)\pi}{n}\bigg)\exp\bigg(i\frac{\pi}{n}\bigg)+\sin\bigg(\frac{m\pi}{n}\bigg)\bigg)$$

For $m=n-1$, we found that $\sin\bigg(\dfrac{(m+1)\pi}{n}\bigg)\exp\bigg(i\dfrac{\pi}{n}\bigg)+\sin\bigg(\dfrac{m\pi}{n}\bigg)=\sin\bigg(\dfrac{\pi}{n}\bigg).~$ As

the first factor in $P$ is $\sin\bigg(\dfrac{\pi}{n}\bigg)\exp\bigg(i\dfrac{\pi}{n}\bigg),~$ we finally find that $\overline{P}=-P$, and we are done.