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.

You are trying to maximize the value of $|z^2-(1+2i)|$ where $z^2$ is a point on the circle, centre $0$ radius $25$
As has been pointed out by Terra Hyde, $z^2$ must be the point on the circle on the other side of the origin from $1+2i$ which would be collinear with $1+2i$ and the origin. Therefore 
$$z^2=25(-\cos\theta-i\sin\theta)$$, where $\tan\theta=2$
Therefore $$z^4=625(\cos2\theta+i\sin2\theta)=375+500i$$
