Too Complex to Solve!

Beautiful problem!

To make the distance maximal, we want the two numbers to point in opposite directions in the complex plane: $z^5=-k(1+2i)z^3$ for some positive $k$ . Dividing by $z^3$ we find $z^2=-k(1+2i)$ . We take the modulus on both sides to find $|z|^2=25=k\sqrt{5}$ , so $k=\sqrt{125}$ and $z^2=-\sqrt{125}(1+2i)$ . Finally, $z^4=125(-3+4i)$ , so that $c+d=125(-3+4)=\boxed{125}$

Let z = r*cis(θ) Maximising distance means arguments are pi apart so arg(z^5) = arg(z^3) + arctan(2) + pi therefore using De Moivre's 5θ = 3θ + arctan(2) + π or θ = [π+arctan(2)] /2

z^4 would then = 5^4 cis(4 ([π+arctan(2)] /2)) = 5^4 cis(2π + 2arctan(2)) = 5^4 cis(2arctan(2)) we can't easily find cis(2arctan(2)) so we rewrite 5^4 cis(2arctan(2)) = (5^2 cis(arctan(2))^2

cos(arctan(2)) and sin(arctan(2)) can be easily found by creating a right angled triangle

z^4 = [ 25 ( 1/sqrt(5) ) + 25i ( 2/sqrt(5) ) ]^2 = (sqrt(125) + sqrt(500)i)^2 = 125 + 500i - 500 = -375 + 500i

-375 + 500 = 125

${ z }^{ 5 }={ \left| z \right| }^{ 5 }cis\left( 5\theta \right)$ ${ z }^{ 3 }\times \left( 1+2i \right) ={ \left| z \right| }^{ 3 }cis\left( 3\theta \right) \times \sqrt { 5 } cis\left( \alpha \right) ={ \left| z \right| }^{ 3 }\sqrt { 5 } cis\left( 3\theta +\alpha \right)$

(alpha is known --->cos(alpha)=1/(sqrt5) and sin(alpha)=2/(sqrt5))

As stated by Otto Bretscher: "To make the distance maximal, we want the two numbers to point in opposite directions in the complex plain". That's because these two numbers are just rotations of the original complex z (followed by increase in the absolute value). So to get the maximum distance we need opposite directions. That means:

$5\theta =3\theta +\alpha +\pi \\ or\quad 5\theta +\pi =3\theta +\alpha$

As we want to find z^4, let's find cis(4*theta). Binding the equations above: $4\theta =2\alpha \pm 2\pi \rightarrow sin\left( 4\theta \right) =sin\left( 2\alpha \right) ;cos\left( 4\theta \right) =cos\left( 2\alpha \right)$

We know alpha ---> sin(2 alpha)=4/5 and cos(2 alpha)=-3/5. So, ${ z }^{ 4 }={ \left| z \right| }^{ 4 }cis\left( 4\theta \right) ={ 5 }^{ 4 }\times \left( -\frac { 3 }{ 5 } +\frac { 4 }{ 5 } i \right) =-3\times { 5 }^{ 3 }+4\times { 5 }^{ 3 }i$

Therefore, the answer is 5^3= $\boxed{125}$

$|z^5-z^3(1+2i)|=|z^3||z^2-1-2i|$

since $z^2$ lives on the circle of radius 25 and we need to keep $z^2$ as far away from $1+2i$ as possible, we immediately get

$z^2=-5\sqrt{5}(1+2i)$

and

$z^4=125(-3+4i)$