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.

Let $\theta$ denote the argument of $z$, then the arguments of $(1+2i)z^3$ and $z^5$ are respectively $3\theta +\arctan(2)$ and $5\theta$. Now imagine that these points lie on concentric circles, so in order to maximize the distance, the difference between the arguments must be $\pi$ (or $-\pi$ it won't matter): 
$$5\theta - 3\theta -\arctan(2)=\pi$$
$$\tan(5\theta - 3\theta -\arctan(2))=0$$
$$\tan(2\theta -\arctan(2))=0=\frac{\tan(2\theta)-2}{1+2\tan(2\theta)}$$
$$\tan(2\theta)=2$$
$$\theta=\frac{1}{2}\arctan(2)\tag{A}$$
$$\tan(\theta)=\frac{1}{2}(\sqrt{5}-1)=\frac{b}{a}$$
$$b^2=\frac{1}{2}(3-\sqrt{5})a^2$$
$$a^2+b^2=25$$
$$a^2=\frac{25+5\sqrt{5}}{2}$$
$$b^2=\frac{25-5\sqrt{5}}{2}$$
ALSO, to get c and d directly, multiply both sides of $(\text{A})$ by 4:
$$4\theta=2\arctan(2)$$
$$tan(4\theta)=-\frac{4}{3}=\frac{d}{c}$$
and we have that
$$\sqrt{c^2+d^2}=5^4$$
Finally, solve for $c$ and $d$.
