Solve a complex number equation I am start learning complex number in my university. I don't know a lot about complex number but I know most of the facts and rules about them.
I have the equation:
$$z^{10} + (z - 1)^{10} = 0$$
It has $10$ roots and they are $z_1$, $\overline{z_1}$, $z_2$, $\overline{z_2}$,..., $z_5$, $\overline{z_5}$. We want to find value of:
$$\sum_{i=1}^5 {\frac{1}{z_i \overline{z_i}}}$$
another method for simplification:   
$\frac{1}{z_k} = 1 - e^{i\frac{\pi(2k-1)}{10}}$  
we get modulus from both side:
$\frac{1}{|z_k|}=|1-\cos(\frac{\pi(2k-1)}{10}) - i\sin(\frac{\pi(2k-1)}{10}) =$
$1 - (1 - 2\sin(\frac{\pi(2k-1)}{20})^{2}-2i\sin(\frac{\pi(2k-1)}{20})\cos(\frac{\pi(2k-1)}{20})|=$
$2|\sin(\frac{\pi(2k-1)}{20})||\sin(\frac{\pi(2k-1)}{20})-i\cos(\frac{\pi(2k-1)}{20})| =$
for $1 \le k \le 5$:
$\frac{1}{|z_k|} = 2\sin(\frac{\pi(2k-1)}{20})$
Is it possible to help me solve this?  Thanks everyone.


We will show
  $$\sum_{k=1}^5 {\frac{1}{z_k \overline{z_k}}}=10.$$

Note that $z^{10} + (z - 1)^{10} = 0$ is equivalent to
$$\left(1-\frac{1}{z}\right)^{10}=-1=e^{-i\pi}.$$
Hence for $k=1,2,3,4,5$, 
$$1-\frac{1}{z_k}=e^{i\pi(2k-1)/10}$$
which implies 
$$\frac{1}{z_k \overline{z_k}}=(1-e^{i\pi(2k-1)/10})(1-e^{-i\pi(2k-1)/10})=2-2\cos\left(\frac{\pi(2k-1)}{10}\right).$$
Therefore 
$$\sum_{k=1}^5 {\frac{1}{z_k \overline{z_k}}}=\sum_{k=1}^5\left(2-2\cos\left(\frac{\pi(2k-1)}{10}\right)\right)=10-2\sum_{k=1}^5\cos\left(\frac{\pi(2k-1)}{10}\right).$$
So it remains to show that the following sum is zero
$$\sum_{k=1}^5\cos\left(\frac{\pi(2k-1)}{10}\right)=\cos\left(\frac{\pi}{10}\right)+\cos\left(\frac{3\pi}{10}\right)+\cos\left(\frac{5\pi}{10}\right)+\cos\left(\frac{7\pi}{10}\right)+\cos\left(\frac{9\pi}{10}\right)$$
which holds by the symmetry $\cos(\pi-x)=-\cos(x)$.
