My question is that how can we evaluate
$\tan^2\left(\dfrac{\pi}{16} \right)+\tan^2\left(\dfrac{3\pi}{16} \right)+\tan^2\left(\dfrac{5\pi}{16} \right)+\tan^2\left(\dfrac{7\pi}{16} \right)$
by using complex numbers. I know how to do it using properties of trigonometry but my professor ask us to use complex numbers to solve this problem.
Consider the complex number equation:
$$(z-1)^8+(z+1)^8=0 \tag{1}$$
$$\iff \left(\frac{1+z}{1-z} \right)^8=-1 \tag{2}$$
The roots of $(2)$ are $$z=i\tan\left(\frac{2k+1}{16} \right) \tag{$k=-4,-3,-2,\cdots, 3$}$$
i.e. $$z=\pm i\tan \left(\frac{\pi}{16} \right), \pm i\tan \left(\frac{3\pi}{16} \right), \pm i\tan \left(\frac{5\pi}{16} \right) \mathrm{and} \pm i\tan \left(\frac{7\pi}{16} \right)$$
Putting back to $(1)$, we have
\begin{align} (z-1)^8+(z+1)^8 &=2\prod_{k=0}^3\left(z-i\tan\left(\frac{2k+1}{16} \right)\pi \right)\left(z+i\tan\left(\frac{2k+1}{16} \right)\pi \right)\\ &=2\prod_{k=0}^3\left(z^2+\tan^2\left(\frac{2k+1}{16} \right)\pi \right) \end{align}
Comparing coefficient of $z^6$, we have
$$\tan^2 \left(\frac{\pi}{16} \right)+\tan^2 \left(\frac{3\pi}{16} \right)+\tan^2 \left(\frac{5\pi}{16} \right)+\tan^2 \left(\frac{7\pi}{16} \right)=28$$
