The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon? The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?
Solution:
Looking at the coefficients, we are immediately reminded of the binomial expansion of ${\left(x+1\right)}^{4}$.
Modifying this slightly, we can write the given equation as ${\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}$
$\star$
We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: $z^{4}=2^{\frac{1}{2}}$
Because the roots of this equation are created by rotating $\frac{\pi}{2}$ radians successively about the origin, the quadrilateral is a square.
We know that half the diagonal length of the square is ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$
Therefore, the area of the square is $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}}$
After the $\star$ I become completely lost. "We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: $z^{4}=2^{\frac{1}{2}}$" Can you please show me exactly how this is achieved, perhaps visually? I am especially confused about what in the equation gets edited to obtain a $- \pi/ 4$ rotation.

If we translate (preserves area) to the new variable $w=z+i$ we get $w^4=1+i$. Solutions are vertices of a square with side length $2^{5/8}$. Hence the area would be $2^{5/4}$.
