# proof of radius of convergence of a complex function

Without loss of generality, it may be assumed that $z_{0}=0$.

Let $c_{n}$ denote the coefficient of the $n$-th term in the Taylor series of $f$ about $0$. Let $r$ be a real number such that $0. Then $c_{n}$ may be expressed as an integral using the Cauchy integral formula.

 $c_{n}={1\over 2\pi i}\oint_{|z|=r}{f(z)\over z^{n+1}}\,dz={1\over 2\pi r^{n}}% \int_{-\pi}^{+\pi}e^{-n\theta}f(re^{i\theta})\,d\theta$

Since $f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $|f| for some constant $B>0$ on the circle $|z|=r$. Hence, we have

 $|c_{n}|={1\over 2\pi r^{n}}\left|\int_{-\pi}^{+\pi}e^{-n\theta}f(re^{i\theta})% \,d\theta\right|\leq{1\over 2\pi r^{n}}\int_{-\pi}^{+\pi}|e^{-n\theta}f(re^{i% \theta})|\,d\theta\leq{1\over 2\pi r^{n}}\int_{-\pi}^{+\pi}Bd\theta={B\over r^% {n}}$

Consequently, $\sqrt[n]{c_{n}}\leq\sqrt[n]{B}/r$. Since $\lim_{n\to\infty}\sqrt[n]{B}=1$, the radius of convergence must be greater than or equal to $r$. Since this is true for all $r, it follows that the radius of convergence is greater than or equal to $R$.

Title proof of radius of convergence of a complex function ProofOfRadiusOfConvergenceOfAComplexFunction 2013-03-22 14:40:35 2013-03-22 14:40:35 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Proof msc 30B10