Answer: 540

Small Hint:

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta)$ and take the real part of each term

Big Hint:

You get an expression of the form $a\cos\theta + b\sin\theta,$ whose maximum is $\sqrt{a^2 + b^2}$

Solution:

Write $z = 4(\cos\theta + \mathrm{i}\sin\theta),$ so $\frac{1}{z} = \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta).$ The real part of $(75 + 117\mathrm{i})z$ is $4(75\cos\theta - 117\sin\theta)$ $= 300\cos\theta - 468\sin\theta,$ and the real part of $(96 + 144\mathrm{i}) \cdot \frac{1}{4}(\cos\theta - \mathrm{i}\sin\theta)$ is $24\cos\theta + 36\sin\theta.$

The total real part is $324\cos\theta - 432\sin\theta,$ whose maximum over $\theta$ is $$\begin{aligned} &\sqrt{324^2 + 432^2} \\ &= 108\sqrt{3^2 + 4^2} \\ &= 108 \cdot 5 = 540. \end{aligned}$$