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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.

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I think the description "polynomial of degree $3/2$" for this bound is rather unusual. – joriki Nov 2 '12 at 1:32
Let $f(z) = a_0 + a_1z + a_2z^2 + \cdots$ be the Taylor expansion of $f$ around the origin. To show $f$ is a linear polynomial, you must show $a_2 = 0$, $a_3 = 0$, etc. What do the Cauchy estimates say about the size of these coefficients? – froggie Nov 2 '12 at 1:36

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, $$f(z) = f(0) + f'(0) z$$

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Instead of using Cauchy estimate as shown in the comments, alternatively you may apply the maximum modulus principle to the function $\frac{f(z)-f(0)-f'(0)z}{z^2}$ to show that it is constantly $0$.

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