Liouville and another beautiful year

From $f(2016) = 6048 + 8064i$ $\implies f(z) = 3z+4zi$ . Therefore,

$\begin{aligned} f(3-4i) & = 3(3-4i) + 4(3-4i)i \\ & = 9-12i + 12i -16i^2 \\ & = 9 + 16 \\ & = \boxed{25} \end{aligned}$

Since $f(z)$ is entire, it has a Taylor series expansion $f(z) \; = \; \sum_{n=0}^\infty a_nz^n \qquad z \in \mathbb{C}$ where $a_n \; = \; \frac{1}{2\pi i}\int_{|z|=R} \frac{f(x)}{z^{n+1}}\,dz \qquad n \ge 0$ where this formula for $a_n$ works for any $R > 0$ . In this case we see that $|a_n| \; \le \; \frac{1}{2\pi}\int_0^{2\pi} \frac{|f(Re^{i\theta})|}{R^{n+1}}\,Rd\theta \; \le \; \frac{2016}{R^{n-1}} \qquad n \ge 0$ Letting $n \to \infty$ we deduce that $a_n = 0$ for all $n \ge 2$ , and hence that $f(z)$ is a linear polynomial. Since $f(0) = 0$ we deduce that $f(z) = kz$ for some constant $k$ . Fitting the information about $f(2016)$ , we see that $k = 3+4i$ . Thus $f(3-4i) = 3^2 + 4^2 =\boxed{25}$ .