Answer to Question #220473 in Complex Analysis for usc

Question

the roots of the equation z^4+az^3+(12+9i)az^2+bz=0( where a nd b are complex numbers) are the vertices of a square

Answer 1

Let $c$ be the center of the square, and $r$ be the distance from the center to vertices.

Then the roots of the given polynomial can be written as $c+re^{i\theta}, c+ire^{i\theta}, c-re^{i\theta}, c-ire^{i\theta}$ for some $\theta\in[-\pi,\pi)$. Therefore the given polynomial must be equal to the product

$(z-(c+re^{i\theta})) (z-(c+ire^{i\theta}))(z-(c-re^{i\theta}))(z-(c-ire^{i\theta}))=$

$((z-c)-re^{i\theta})( (z-c)+re^{i\theta})((z-c)-ire^{i\theta})((z-c)+ire^{i\theta}))=$

$((z-c)^2-r^2e^{2i\theta})((z-c)^2+r^2e^{2i\theta})=(z-c)^4-r^4e^{4i\theta}=$

$z^4-4cz^3+6c^2z^2-4cz+c^4-r^4e^{4i\theta}$

Comparing the coefficients, we obtain

$a=b=-4c$, $c^4=r^4e^{4i\theta}$ and $6c^2=(12+9i)a=-4c(12+9i)$.

From the last equation either c=0 or $c=-8-6i$.

The case c=0 is degenerated, since in this case r=0 and the square is, in fact, a point.

We do not consider this case. Therefore, $c=-8-6i$, $r=|c|=\sqrt{(-8)^2+(-6)^2}=10$

$e^{i\theta}=\frac{c}{r}e^{\frac{2\pi i}{4}k}=(-\frac{4}{5}-\frac{3}{5}i)e^{\frac{\pi ik}{2}}$ with k=0,1,2 and 3 (each value of k is suitable).

$a=b=-4c=-4(-8-6i)=32+24i$

Answer. $a=b=32+24i$.