Q: Entire Complex Functions and Cauchy Integral Formula Suppose that f(z) is an entire function.  Show that $\exists c\in \mathbb{R}, c>0$ such that $|f(z)|\leq c|z| \forall z\in \mathbb{C} \Longrightarrow f(z)=az, a\in \mathbb{C}$ and $|a|\leq c$.
Hint: use the formula:
$f''(z)=\frac{2!}{2\pi i}\int_{|\zeta|=R}\frac{f(\zeta)}{(\zeta-z)^3}d\zeta$ to estimate $|f''(z)|$ when R goes to positive infinity.
I'm pretty lost, so some insight would be appreciated.  I also have no idea how the hint relates to the question...

A: All you need to do is check that
$$
|f''(z)| \leq \frac{1}{\pi} \int_{|\zeta| = R}c\frac{|\zeta|}{|\zeta - z|^3}d\zeta \to 0
$$
as $R \to \infty$. Hence, $f'' \equiv 0$ which gives you what you want.

A: First note that 
$$
|f^{(n)}(z)| \leq \frac{(n!)}{2\pi} \int_{|\zeta| = R}c\frac{|\zeta|}{|\zeta - z|^{n+1}}d\zeta \to 0
$$
for all $n\ge 2$. This means that $f$ is linear polynomial, i.e., $f(z)=az+b$ for some $a,b\in\mathbb{C}$.
For $z=0$ you have $|b|\leq 0\iff b=0$. Consequently, we have $|a||z|\leq c|z|,\forall z\in\mathbb{C}$ that implies $|a|\le c$ as you wanted.
