Q:

Finding the value of this complex integral

Suppose that $f$ is an entire function such that $|f(z)| \leq M|z|$ for some real $M$. Show that $f(z) = az + b$, where $a,b$ are complex numbers.
My initial thought was to show that $f''(z) = 0$. To start, I used the Cauchy Inequality formula in the following way:
$$ |f''(z)|≤\frac{1}{\pi} \int_{|ζ|=R} \frac{M|ζ|}{|ζ−z|^3}dζ $$
If I show that the above integral goes to $0$ as $R$ approaches $\infty$, the proof is complete. However, showing this is exactly where I'm stuck. What am I missing here?

A:

Suppose $f$ is an entire function such that $|f(z)|\leq M|z|$ for some real constant $M$.
Then, by Cauchy's Integral Formula,
$\displaystyle f^{\prime}(z) = \frac{1!}{2\pi i} \int_{\gamma_{\rho}} \frac{f(\zeta)}{(\zeta - z)^{2}}d\zeta$.
So, $\displaystyle |f^{\prime}(z)| = \frac{1}{2\pi }\left\vert \int_{\gamma_{\rho}} \frac{f(\zeta)}{(\zeta - z)^{2}}d\zeta\right\vert \leq \frac{1}{2\pi} \int_{\gamma_{\rho}}\left\vert\frac{f(\zeta)}{(\zeta - z)^{2}}d \zeta\right\vert = \frac{1}{2\pi} \int_{\gamma_{\rho}}\frac{|f(\zeta)|}{|(\zeta - z)^{2}|}|d\zeta| \leq \frac{1}{2\pi}\int_{\gamma_{\rho}}\frac{M|\zeta|}{|(\zeta-z)^{2}|}|d\zeta| = \frac{M}{2\pi}\int_{\gamma_{\rho}}\frac{|\zeta|}{|(\zeta-z)^{2}|}|d\zeta| = \frac{MI}{2\pi}$, where $\displaystyle I = \int_{\gamma_{\rho}}\frac{|\zeta|}{|(\zeta-z)^{2}|}|d\zeta|$.
Thus, we have established that $f^{\prime}(z)$ is bounded. It is certainly differentiable, since $f$ itself being first-differentiable implies that $f$ is infinitely differentiable.
By Liouville's Theorem, then, $f^{\prime}(z)$ is constant. I.e., $f^{\prime}(z) = a$, for $a \in \mathbb{C}$.
Integrating this, we obtain that $f(z) = \int f^{\prime}(z) dz = \int a dz = az+b$, where $b \in \mathbb{C}$ is a constant of integration.

