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 $0 < r < R$ Then $c_n$ may be expressed as an integral using the Cauchy integral formula. $$c_n = {1 \over 2 \pi i} \oint_{|z| = r} {f(z) \over z^{n+1}} \, dz = {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta$$

Since $f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $|f| < B$ for some constant $B > 0$ on the circle $|z| = r$ Hence, we have $$|c_n| = {1 \over 2 \pi r^n} \left| \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta \right| \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} | e^{-n \theta} f(r e^{i \theta}) | \, d \theta \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} B d \theta = {B \over r^n}$$

Consequently, $\sqrt[n]{c_n} \le \sqrt[n]{B} / r$ 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 $r < R$ it follows that the radius of convergence is greater than or equal to $R$