Good afternoon. I have the following problem.
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. Find the area of the polygon.
So its been a bit since I had Complex, but if I remembered correctly, the roots of a complex number are distributed equidistantly around the origin in the form of a regular polygon and by the FT of A, there must be 4 of these roots and thus the area is a square. So if I can find two adjacent roots, their distance squared must be the area. Now, the polynomial equation can be written as
$$z^4+4z^3i-6z^2-4zi-i=(z+i)^4-1-i=0 \quad \rightarrow \quad (z+i)^4=1+i=\sqrt{2}e^{(\frac{i\pi}{4}+2k\pi)}$$
Now taking a 4th root gives
$$z+i=\sqrt[8]{2}e^{(\frac{i\pi}{16}+\frac{k\pi}{4})}$$
if we let $w=z+1$, then the +1 just shifts the vertices of the polygon left by 1 unit and doesn't change the area it, so we don't need to worry about the +1 and we can just solve for $z$. The 4 solutions then are
$$2^{1/8}e^{i\pi/16}, 2^{1/8}e^{9i\pi/16}, 2^{1/8}e^{17i\pi/16}, 2^{1/8}e^{26i\pi/16}$$
Looking at the first two as consecutive solutions (they would be adjacent vertices), we can write them as
$$2^{1/8}\left(\cos{\frac{\pi}{16}+i\sin{\frac{\pi}{16}}}\right) \quad 2^{1/8}\left(\cos{\frac{9\pi}{16}+i\sin{\frac{9\pi}{16}}}\right)$$
And then the area of the square must be
$$A=2^{1/4}\left[\left(\cos{\frac{9\pi}{16}}-\cos{\frac{\pi}{16}}\right)^2+\left(\sin{\frac{9\pi}{16}}-\sin
{\frac{\pi}{16}}\right)^2\right]$$
Expanding out gives
$$A=2^{1/4}\left(2-2\cos{\frac{9\pi}{16}}\cos{\frac{\pi}{16}}-2\sin{\frac{9\pi}{16}}\sin{\frac{\pi}{16}}\right)=2^{1/4}\left(2-2\cos{\frac{\pi}{2}}\right)=2^{5/4}$$
First off: Is this correct. It's not HW, I'm not a student. Second, is there a quicker way? I checked a couple of problems like this one here, but I think because it's ultimately reducible to a binomial power that we can make the regular polygon assumption.
Best Answer
Your computation is correct.
Alternatively you can argue that the polygon is a square, inscribed in a circle of radius $r = |1+i|^{1/4} = 2^{1/8}$. The side length of the square is $l = \sqrt 2 r$ and its area is $A = l^2 = 2r^2 = 2^{5/4}$.