<s>[INST] find the complex number $z^4$ 
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
  Find $c+d$.

I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance between those two points. 
$|z^2| = 25$ means since: $z^2 = a^2 - b^2 + 2abi$ that:
$625 = (a^2 - b^2)^2 + 4a^2b^2 = a^4 + b^4 + 2a^2b^2$
But that doesnt help much.
 [/INST] If $z=5e^{i\theta}$, and we use $1+2i=\sqrt5e^{i\phi}=z_0$, then we have
$$|z^5-z^3z_0|=125|z^2-z_0|$$
as the thing to maximise, but this implies we need $z^2$ on the diameter through $z_0$, so we need $\theta=\frac{\phi+\pi}{2}$.
Then, $$z^4=625e^{4i\theta}=625e^{2i\phi+2i\pi}=625e^{2i\phi}=625\cos(2\phi)+625i\sin(2\phi)$$
Knowing that $\phi=\arctan(2)$ Provides a direct route to the numerical answer.
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