Let $z^{2n}=1$
$$\implies z^{2n}-1=(z-z_1)(z-z_2)(z-z_3)....(z-z_{2n})$$
where $z_1,z_2...z_{2n}$ are the roots of unity
Since $z_k^{2n}=e^{2k\pi i}$,
$$z_k=e^{i(\frac{k\pi}{n})}=\cos\bigg(\frac{k\pi}{n}\bigg)+i\sin\bigg(\frac{k\pi}{n}\bigg)$$
Notice that
$$z_{2n-k}=\cos\bigg(\frac{(2n-k)\pi}{n}\bigg)+i\sin\bigg(\frac{(2n-k)\pi}{n}\bigg)$$
$$=\cos\bigg(\frac{2n\pi-k\pi}{n}\bigg)+i\sin\bigg(\frac{2n\pi-k\pi}{n}\bigg)$$
$$=\cos\bigg(2\pi-\frac{k\pi}{n}\bigg)+i\sin\bigg(2\pi-\frac{k\pi}{n}\bigg)$$
$$\cos\bigg(\frac{k\pi}{n}\bigg)-i\sin\bigg(\frac{k\pi}{n}\bigg)=\bar{z_k}$$
Also,
$$z_n=\cos(\pi)+i\sin(\pi)=-1$$
and
$$z_{2n}=\cos(2\pi)+i\sin(2\pi)=1$$
Grouping all the conjugates together and then multiplying them,
$$z^{2n}-1=(z-z_1)(z-z_{2n-1})(z-z_2)(z-z_{2n-2})....(z_n)(z_{2n-n})$$
$$=(z-z_1)(z-\bar{z_1})(z-z_2)(z-\bar{z_2})....(z-1)(z+1)$$
$$=\big[z^2-z(z_1+\bar{z_1})+|z_1^2|\big]\big[z^2-z(z_2+\bar{z_2})+|z_2^2|\big]....(z^2-1)$$
Since $|z_k|^2=1$ and $(z_k+\bar{z_k})=2Re(z_k)=2\cos(\frac{k\pi}{n})$,
$$z^{2n}-1=\big[z^2+1-2z\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z^2+1-2z\cos\bigg(\frac{2\pi}{n}\bigg)\big]....(z^2-1)$$
$$\implies \frac{z^{2n}-1}{z^2-1}=\big[z^2+1-2z\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z^2+1-2z\cos\bigg(\frac{2\pi}{n}\bigg)\big]....$$
Dividing both sides by $z^{n-1}$,
$$\frac{z^n-\frac{1}{z^n}}{z-\frac{1}{z}}=\big[z+\frac{1}{z}-2\cos\bigg(\frac{\pi}{n}\bigg)\big]\big[z+\frac{1}{z}-2\cos\bigg(\frac{2\pi}{n}\bigg)\big]....$$
Substituting $z=\cos(\theta)+i\sin(\theta)$
$\implies \frac{1}{z}=\cos(\theta)-i\sin(\theta)$
Also, $z^n=\cos(n\theta)+i\sin(n\theta)$ $\implies\frac{1}{z^n}=\cos(n\theta)-i\sin(n\theta)$
$$\implies\frac{z^n-\frac{1}{z^n}}{z-\frac{1}{z}}= \frac{2i\sin(n\theta)}{2i\sin(\theta)}=\bigg[2\cos(\theta)-2\cos\bigg(\frac{\pi}{n}\bigg)\bigg]\bigg[2\cos(\theta)-2\cos\bigg(\frac{2\pi}{n}\bigg)\bigg].....$$
$$\implies 2^{n-1}\prod_{k=1}^{n-1}[\cos(\theta)-\cos\big(\frac{k\pi}{n}\big)]=\fbox{$\frac{\sin(n\theta)}{\sin(\theta)}$}$$
Best Answer
For the first, it is equal to $i^{-i}.$ So, the log is equal to $-i(\pi i/2 + 2ki\pi) = \pi/2 +2 k \pi.$
The second, before you square, you have the real part of $x=\omega + \omega^2 + \omega^4,$ where $\omega$ is the primitive seventh root of unity. Notice that the conjugate of this expression is $\omega^6 + \omega^5 + \omega^3 = 1-x.$ Since the real part of $x$ is the same as that of $\overline{x},$ we have that the real part of $x$ is $1/2,$ so its square is $1/4.$