Extend $f(z)=\frac{1}{z^n +z^{n-1}+...+z^2 + z^{-n}}+\frac{c}{z-1}$

Question

find $c $ such that $ f(z)=\frac{1}{z^n +z^{n-1}+...+z^2 + z^{-n}}+\frac{c}{z-1}$ can be extended to be analytic at $z=1$ , when $n\in \mathbb{N}$ when $n$ is fixed.

The given function I write it as $f(z)=\frac{z^n(z-1)+c(z^n+1)(z^{n+1} -1) }{(z-1)(z^n +1) (z^{n+1} + 1)}$

Further i tried to evaluate limit at 1, so that I can choose c , so that my limit will always exist...

I dont know my approach is false.. Help me please

Answer 1

Hint

Surely $f(x)$ is analytic at a small enough neiborhood of $z=1$, therefore $z=1$ is an isolated singularity of $f(x)$ for all values of $c$ EXCEPT ONE for which:$$\lim_{z\to 1}(z-1)f(z)=0$$Now what's the value of $c$?