Entire function bounded by polynomial of degree 3/2 must be linear. The problem is as follows:

Suppose $f$ entire satisfying
  $$ |f(z)| \leq A + B |z|^{3/2} $$
  for some fixed $A,B > 0$. Prove that $f$ is a linear polynomial.

I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how.

The first technique that comes to mind is the Cauchy integral formula.
$$f^{(n)}(0) = \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz$$
Hence, we have that
$$\left \vert f^{(n)}(0) \right \vert = \left \vert \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz \right \vert \leq \oint_{\Gamma} \dfrac{\left \vert f(z) \right \vert}{\left \vert z \right \vert^{n+1}} dz \leq \oint_{\Gamma} \dfrac{A + B \vert z \vert^{3/2}}{\left \vert z \right \vert^{n+1}} dz$$
Take $\Gamma$ to be a circle of a very large radius $R$. We then have that$$\left \vert f^{(n)}(0) \right \vert \leq \dfrac{A_1}{R^n} + \dfrac{B_1}{R^{n-3/2}}$$
We can let $R$ to be arbitrarily large and hence for $n \geq 2$, we have that
$$f^{(n)}(0) = 0 \,\,\,\, \forall n \geq 2$$
Hence, $$f(z) = f(0) + f'(0) z$$
