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