Proof. Taking \(C\) to be a circle of arbitrary radius say \(R\) on a point \(z_0\). It follows that \[f'(z_0) = \frac {1}{2\pi i}\int _C \frac {f(z)}{\big (z - z_0\big )^2}dz\] By hypothesis, if \(f(z)\) is bounded for all \(z\in \mathbb {C}\).
Then so is \(\left |f(z)\right |\), and thus there exist a positive constant \(M\) such that \(\,\left |f(z)\right |< M\,\) for all \(z\).

Setting \(\, z - z_0 = Re^{i\theta }\,,\quad 0\leq \theta \leq 2\pi \,, \quad dz = iRe^{i\theta }d\theta \) \begin {align*} f'(z_0) & = \frac {1}{2\pi i}\int ^{2\pi }_0 \frac {f(z)\cdot iRe^{i\theta } d\theta }{R^2e^{2i\theta }}\\\\ & = \frac {1}{2\pi R}\int ^{2\pi }_0\frac {f(z)d\theta }{e^{i\theta }} \end {align*}

Take modulus on both sides \begin {align*} \left |f'(z_0)\right | & = \left |\frac {1}{2\pi R}\displaystyle {\int ^{2\pi }_0}\frac {f(z)d\theta }{e^{i\theta }}\right |\\\\ & \leq \frac {1}{2\pi R}\int ^{2\pi }_0\left |\frac {f(z)}{e^{i\theta }}\right |d\theta \\\\ & \leq \frac {M}{2\pi R}\int ^{2\pi }_0 d\theta \\\\ & = \frac {M}{R} \end {align*}

Thus \(\,\left |f'(z_0)\right |\leq \frac {M}{R}\,,\,\) Since \(R\) is arbitrary, i.e the inequality is true for all \(R>0\).

Take \(R\) arbitrary large \(\, ( R\longrightarrow \infty )\) \[0 \leq \left |f'(z_0)\right | \leq \frac {M}{R}\] Then \(\,\left |f'(z_0)\right | =0 \,\) for all \(z_0\) in the complex plane. Thus, this is only possible if \(f(z)\) is identically constant. i.e \(\, f(z) = \) constant.

This implies that if a non constant function is analytic in the entire plane then it cannot be bounded. □