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 . Then $c_n$ may be expressed as an integral using the Cauchy integral formula.

$$c_n=\frac{1}{2\pi i}{\oint }_{|z|=r}\frac{f(z)}{z^{n+1}}𝑑z=\frac{1}{2\pi r^n}{\int }_{-\pi }^{+\pi }{e}^{-n\theta }f(r{e}^{i\theta })𝑑\theta$$

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

$$|c_n|=\frac{1}{2\pi r^n}\left|{\int }_{-\pi }^{+\pi }{e}^{-n\theta }f(r{e}^{i\theta })𝑑\theta \right|\le \frac{1}{2\pi r^n}{\int }_{-\pi }^{+\pi }|e^{-n\theta }f(r{e}^{i\theta })|𝑑\theta \le \frac{1}{2\pi r^n}{\int }_{-\pi }^{+\pi }B𝑑\theta =\frac{B}{r^n}$$

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