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

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

• Are you familiar with Riemann's theorem en.wikipedia.org/wiki/Removable_singularity? – copper.hat Mar 27 at 13:30
• No. It is the first time i am encounter with such problem.. Let me check what is it about. – user485546 Mar 27 at 13:32
• Actually, it is not clear how your $f$ is defined. There is an unexpected $z^2$ in the denominator of the first term of $f$. If that is what you intended, you need to give a more explicit description of $f$. – copper.hat Mar 27 at 13:38
• This is an assignment question...!! No changes – user485546 Mar 27 at 13:42
• It doesn't matter, the form of $f$ as you have it above is undefined. What is the term immediate;y preceding $z^2$ in the denominator? – copper.hat Mar 27 at 13:45

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 $$c$$?