Solve $z^5=-32$ and draw its solutions in complex space, then describe their characteristic geometrical property.

Question

I'm solving past exam questions in preparation for an Applied Mathematics course. I came to the following exercise, which poses some difficulty. If it's any indication of difficulty, the exercise is only Part 3-A of the sheet, graded for 10%

Solve the equation $z^5=-32$ and draw its solutions in complex space, then describe their characteristic geometrical property.

Is it asking to convert z to polar form, then use DeMoivres theorem as I've seen in solutions around the net? If that is the case, how can I work out the $\theta$ angle to be used?

Additionally, what does it refer to as its characteristic geometrical property?

Any answers would be extremely appreciated as they'd help to get me out of the ditch. I'm completely stalled.

Answer 1

Let $z=\rho e^{i\theta}$. Representing $-32$ in the Argand plane you see that, in polar form, it is $-32=32e^{i\pi}$, so your equation becomes: $$ \left(\rho e^{i\theta}\right)^5=32e^{i\pi} $$ that gives: $$ \rho e^{i\theta}=\left( 32e^{i\pi}\right)^{1/5}=2\left(e^{i\pi}\right)^{1/5} $$ Now, since $e^{i\pi}=e^{i(\pi+2k\pi)}$ we have five distinct values for $\left(e^{i\pi}\right)^{1/5}$:

$$ \left(e^{i(\pi+2k\pi)}\right)^{1/5}=e^{i\pi/5+2ik\pi/5} $$ for $k=0,1,2,3,4$ and the five complex roots of $-32$ are: $$ z_1=2e^{i\pi/5} \quad z_2=2e^{3i\pi/5} \quad z_3=2e^{5i\pi/5}=2e^{i\pi}=-2\quad z_4=2e^{7i\pi/5} \quad z_5=2e^{9i\pi/5} $$

In the Argand plane these points are the vertex of a regular pentagon inscribed in a circle of radius $2$ an with a vertex on $-2$.

Answer 2

The solutions should just be $2e^{\frac{2\pi ki}{10}}$, for odd integers $0 \leq k <10$. They are the divisions of the circle of radius 2 centered at the origin, divided into 5 pieces, rotated about the origin $\frac{2 \pi}{10}$ in the complex plane.

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Here's more details:

$z^5=-32 \implies 32((\frac{z}{2})^5+1)=0$. Setting $y=z/2$, and noting that $y^{10}-1=0 \implies (y^5-1)(y^5+1)=0$ indicates that the values of $y$ are the $10$th roots of unity for which $y^5=-1$. So $y=e^{\frac{2\pi ki}{10}}$ for odd $k$ (even values of $k$ would be roots of the first factor instead).