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.
 A: I certainly find the description from $\star$ onwards a little confusing, mainly because it's not stated exactly what it is that's supposed to be being translated and rotated.
Ignore the geometry of situation for the moment, and just look at a particular algebraic method of solving the equation, by setting $\ w=z+i\ $ and $\  v=we^\frac{-i\pi}{16}\ $.  From the original equation we get $\ w^4=\sqrt{2}e^\frac{i\pi}{4}\ $ , and then $\ v^4=w^4e^\frac{-i\pi}{4}=\sqrt{2}\ $. The four solutions of this equation for $\ v\ $ are $\ v_1=\sqrt[8]{2}, v_2=\sqrt[8]{2}
\,i, v_3=-\sqrt[8]{2}\ $, and $\ v_4=-\sqrt[8]{2}
\,i\ $, which are vertices of a square, lying on the positive and negative real and imaginary axes at a distance of $\ \sqrt[8]{2}\ $ from the origin.  To recover the solutions for $\ z\ $, you first have to rotate those for $\ v\ $ by $\ \frac{\pi}{16}\ $ to get those for $\ w\ $:
$$
w_1=\sqrt[8]{2}\, e^\frac{i\pi}{16}, w_2=\sqrt[8]{2}\,i\,e^\frac{i\pi}{16}, w_3=-\sqrt[8]{2}\, e^\frac{i\pi}{16}\ ,\text{ and }\ w_4=-\sqrt[8]{2}\,i\ e^\frac{i\pi}{16}\ ,
$$
and then translate these by $\ {-}i\ $ to get those for $\ z\ $:
\begin{align}
z_1&=\sqrt[8]{2}\, e^\frac{i\pi}{16}-i\\
z_2&=\sqrt[8]{2} \,i\, e^\frac{i\pi}{16}-i\\
z_3&=-\sqrt[8]{2}\, e^\frac{i\pi}{16}-i\ ,\text{ and }\\
z_4&=-\sqrt[8]{2}\,i\, e^\frac{i\pi}{16}-i\ .
\end{align}
These are the vertices of a square, congruent to that with vertices $\ v_1,v_2,v_3,v_4\ $, but with its centre at the point $\ {-}i\ $ instead of the origin, and its diagonals rotated $\ \frac{\pi}{16}\ $ anticlockwise from the horizontal and vertical, instead of being exactly horizontal and vertical.
I therefore don't understand why the exposition says "[w]e can apply a translation of $\ {-}i\ $ and a rotation of $\ {-}\frac{\pi}{4}\ $" rather than "[w]e can apply a translation of $\ i\ $ and a rotation of $\ {-}\frac{\pi}{16}\ $".  Unless the author of the exposition has simply made a mistake, he or she must be referring to different operations from the ones I've described above.  But if that's the case, I'm afraid I have no idea what those operations are.
A: 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 $2^{5/4}$.
