Solution: Given that $f$ is an entire function, so it can be represented by its Taylor series expansion around $z = 0 $. Let \begin{equation}\label{eq:25Feb2025-1} f(z) = \sum_{n=0}^{\infty} a_n z^n \end{equation} be the Taylor series of $f$ on an open disc $\vert z \vert < r$, where \[ a_n = \frac{1}{2\pi \iota } \oint_{\vert z \vert = r} \frac{f(w)}{w^{n+1}} \mathrm{d} w = \frac{f^{(n)}(0)}{n!}. \] Since $\vert f(z) \vert \leq A + B \vert z \vert ^{3 / 2}$, on the circle $\vert z \vert = r$, the function is bounded by \[ \vert f(z) \vert \leq A + B r^{3 / 2} \coloneqq M. \] Therefore, using Cauchy's estimate, we get \[ \left\vert a_n \right\vert = \left\vert \frac{f^{(n)}(0)}{n!} \right\vert \leq \left\vert \left( \frac{n!}{r^n} M\right) \times \frac{1}{n!} \right\vert = \frac{M}{r^n}. \] Note that since $A, B$ are fixed real numbers, we can always choose $r>0$ such that $M \leq r^2$. Thus, for $n\geq 2$, \[ \lim_{r \to \infty} \frac{M}{r^n} = 0. \] Therefore, for $n\geq 2$, we have $a_n = 0$. Thus, the Taylor series expansion of $f$ in \eqref{eq:25Feb2025-1} reduces to \[ f(z) = a_0 + a_1 z. \] Thus, $f$ is a linear polynomial.