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.
 A: 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 $$z^4=625(\cos2\theta+i\sin2\theta)=375+500i$$
A: 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.
A: 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 $d$.
