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I have dealt with radius of convergence for simple series, but this one is literally complex:

$\frac{1}{1-z-z^2}=\sum_{n=0}^\infty c_nz^n$

How does one calculate the radius of convergence here? I can't just use the ratio test? Any ideas?

What methods would I use in general? I haven't much experience with complex analysis

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As Micheal pointed, the radius of convergence is just the distance from the origin of the closest singularity, so $\rho=\frac{\sqrt{5}-1}{2}$. You can achieve that also by noticing that: $$ c_n = F_{n+1} = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]\tag{1}$$ since $(1)$ holds for $n\in\{0,1\}$ and $(1-x-x^2)\cdot\frac{1}{1-x-x^2}=1$ implies: $$ \forall n\geq 0,\quad c_{n+2}=c_{n+1}+c_n.\tag{2}$$

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The radius will be the distance from $z=0$ to the smallest root of the denominator. In this case, $1/\phi$.

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  • $\begingroup$ What is $\phi$? Also I like your user picture a heap, can I have it? $\endgroup$ – Funcomplexsum May 31 '15 at 14:34
  • $\begingroup$ No, it's mine :( $\phi$ is the golden ratio, $(\sqrt{5}+1)/2=1.618$ $\endgroup$ – Empy2 May 31 '15 at 14:36
  • $\begingroup$ Please don't use it. $\endgroup$ – Empy2 May 31 '15 at 15:13
  • $\begingroup$ Really? Do you really not want me to use it?? If you are that against it, I feel bad enough to take it down $\endgroup$ – Funcomplexsum May 31 '15 at 15:14
  • $\begingroup$ Thanks, I would appreciate that. $\endgroup$ – Empy2 May 31 '15 at 15:16

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