Let $\{{z_1,..,z_5}\}$ Solutions of $z^5=2i$ Find $x^2=\frac{1}{z_1}+...+\frac{1}{z_5}+i$

Question

Let $\{{z_1,..,z_5}\}$ Solutions of $z^5=2i$

Find $x^2=\frac{1}{z_1}+...+\frac{1}{z_5}+i$

I found the solutions of $z^5=2i$ :

$z_1=\sqrt[5]{2}cis\left(\frac{\pi }{10}\right)$

$z_2=\sqrt[5]{2}cis\left(\frac{\pi }{2}\right)$

$z_3=\sqrt[5]{2}cis\left(\frac{9\pi }{10}\right)$

$z_4=\sqrt[5]{2}cis\left(\frac{13\pi }{10}\right)$

$z_5=\sqrt[5]{2}cis\left(\frac{17\pi }{10}\right)$

now I need to find what is $x^2=\frac{1}{z_1}+...+\frac{1}{z_5}+i$

and here I'm stuck How to continue from here ?

Thanks

Answer 1

The $z_k$'s are the fifth roots of $2i$ and therefore their inverses are the fifth roots of $\frac1{2i}$. So, their sum is $0$ (the sum of all $n$^th roots of a complex number is always $0$ when $n>1$). Therefore, the possible values of $x$ are the square roots of $i$.

Answer 2

Let $w=\frac1z$, then $z^5 =2i$ leads to

$$w^5 + \frac12 i =0$$

which yields

$$ b=w_1+w_2+w_3+ w_4 + w_5=0$$

Thus, $x^2=b+i =i $.