The difficulty is imaginary

How to look at it:

Where the black circle is all possible z, the red line is an arbitrary z, the blue line is z^2, and the green line is z^3.

When you look at it full on, its clear that these lines ends create a triangle at their end points.

$z^3 - z^2$ and $z^3 - z$ make up two of this triangle's lines, the cyan and yellow ones here, respectively.

Come to think of it, the expression we're supposed to maximize looks kind of familiar now...

$\text{area of a triangle} = 1/2 * a * b * \sin(\text{angle between a and b})$

So if the area of this triangle is really similar to our expression we're supposed to maximize its clear that the correct answer is the area of an equilateral triangle with a length of $\sqrt{3}$

But does this comparison check out? Getting rid of the 1/2, we see they certainly have a very similar structure. But how does the sine part translate?

Remember that:

(src: online math learning)

and that the angle between z and z^2, as well as z^3 and z is just the angle of z with regard to the real line.

Remember also that a complex number can be represented by $\text{length }*e^{\text{angle with regard to real line}*i}$ , so the ln of that would be $\text{length}) + \text{angle with regard to real line}*i$

so for our zs, this would simply be the angle multiplied by i.

So the equations really do match up!

"But Jeremy!" I hear you cry, "I'm an abstract mathematician and drawings scare me! Plus, you never actually proved anything regarding the maximum!"

That's okay, there is a more analytical solution. (it does depend on the geometric model, but you can pretend I went through and did all the algebra.)

the expression in the question is equal to $sin(arg(z))*(|z^2||z|+|z^3||z^2|) - sin(2arg(z))|z^3||z|$ , where arg(z) is z's angle with regard to the real line.

this simplifies to $2sin(arg(z)) - sin(2arg(z))$ , taking the derivative and setting it to zero gives us $cos(arg(z))-cos(2arg(z))=0$ which is clearly $\frac{2pi}{3}$

Plugging in the corresponding z = $e^{\frac{2pi}{3}i}$ , we get the answer we're expecting.