Let $w \in \mathbb{C}$ such that $\left|w\right| = 5$, $\operatorname{Im}(w) > 0$, and the distance between $(1 + 2i)w^3$ and $w^5$ is maximum. Find the complex number $w^4$, writing your answer in the form $a + bi$.
Geometrically, $z$ would lie on a circle with radius $5$ centered at the origin. From my understanding, $(1+2i)w^3$ and $w^5$ would also lie on circles, but with different radii.
I was thinking of finding the two points furthest from each other on the circles. However, I'm not sure how to proceed from here. Am I going to involve the distance formula? Any help would be appreciated. Thanks!
$|(1+2i)\omega^3-\omega^5|=|\omega|^3|1+2i-\omega^2|=125|1+2i-\omega^2|$
Note that $\omega^2$ represents a point on the circle centred at $0$ with radius $25$. $|1+2i-\omega^2|$ is the maximum when $\omega^2$ is the point where the circle meets the ray from $1+2i$ to $0$. So,
$$\omega^2=(-1-2i)\cdot\frac{25}{|-1-2i|}$$
$$\omega^4=(-1-2i)^2\cdot\frac{625}{5}=-375+500i$$