Finding the inverse of $\sum\limits_{i=0}^{49}u^i$ where $u$ is a complex root of $x^{50}-2$

Question

Consider $p(x)=x^{50}-2\in\Bbb Q[x]$ and $u\in\Bbb C$ such that $p(u)=0\ \ $ ($u^{50}=2$)

By Eisentsein $p(x)$ is irreducible.

If we define $\mu:=1+u+...+u^{49}$ multiplying by $u$ gives $$u+u^2+...+u^{49}+2=u\cdot\mu=\mu+1$$

$\implies \mu(u-1)-1=0$

How should I use the extended euclidean algorithm to find $\mu^{-1}$?

Answer 1

You have the sum of a geometric sequence, so $$ \sum_{k=0}^{49}u^k=\frac{u^{50}-1}{u-1}=\frac{1}{u-1} $$ because $u^{50}=2$.

The inverse is $u-1$.