Answer: 125

Small Hint:

Factor the distance as $|z|^3 \cdot |1 + 2\mathrm{i} - z^2|,$ so you need the point $z^2$ on the circle of radius $25$ farthest from $1 + 2\mathrm{i}$

Big Hint:

The farthest point lies in exactly the opposite direction: $z^2 = -5\sqrt{5}\,(1 + 2\mathrm{i});$ now square once more

Solution:

The distance is $|(1 + 2\mathrm{i})z^3 - z^5|$ $= |z|^3 \cdot |1 + 2\mathrm{i} - z^2|$ $= 125\,|1 + 2\mathrm{i} - z^2|.$ As $z$ runs over the circle $|z| = 5$ with $b \gt 0,$ the square $z^2$ attains every point of the circle $|w| = 25$ (the condition $b \gt 0$ merely selects one of the two square roots). The point of that circle farthest from $1 + 2\mathrm{i}$ is diametrically opposite in direction: $$\begin{aligned} z^2 &= -25 \cdot \frac{1 + 2\mathrm{i}}{|1 + 2\mathrm{i}|} \\ &= -5\sqrt{5}\,(1 + 2\mathrm{i}). \end{aligned}$$

Squaring, $z^4 = 125\,(1 + 2\mathrm{i})^2$ $= 125\,(-3 + 4\mathrm{i})$ $= -375 + 500\mathrm{i},$ so $c + d = -375 + 500 = 125.$