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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

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  • $\begingroup$ Are you familiar with Riemann's theorem en.wikipedia.org/wiki/Removable_singularity? $\endgroup$ – copper.hat Mar 27 at 13:30
  • $\begingroup$ No. It is the first time i am encounter with such problem.. Let me check what is it about. $\endgroup$ – user485546 Mar 27 at 13:32
  • $\begingroup$ 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 $f$. $\endgroup$ – copper.hat Mar 27 at 13:38
  • $\begingroup$ This is an assignment question...!! No changes $\endgroup$ – user485546 Mar 27 at 13:42
  • $\begingroup$ 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? $\endgroup$ – copper.hat Mar 27 at 13:45
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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 $c$?

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