Visualizing roots of an equation in complex variables

Question

What does it mean to find the roots of an equation involving complex numbers, say z^5-1=0? Is there some way to visualize it?

The way I see it,z would not trace a curved line but a curved plane. How am I to find the roots of a plane?

Can the argand plane instead be visualized as a line?

Answer 1

Consider the more general equation $P(z)=0$, where $P$ is any "reasonable" map.

A polynomial map is OK.

You can construct (or rather ask you favorite computer to construct) the surface defined by the equation $z=\left|P(x+iy)\right|$ and look at its intersection with the plane $z=0$.

** EDIT : **

Density plot is not bad either ...

Here is how, using Maple, for your example ($z^5-1=0$) :

Black corresponds to 0, white to 1 or more.

densityplot(abs((x+I*y)^5-1), x = -1.2 .. 1.2, y = -1.2 .. 1.2, grid = [50, 50], colorstyle = SHADING, style = patchnogrid, scaletorange = 0 .. 1, labels = ["", ""])

Answer 2

In polar coordinates,

$$z^5=r^5e^{i5\theta}=1=e^{0+2k\pi}.$$

Taking the modulus,

$$r^5=1$$ represents the unit circle.

Taking the argument,

$$5\theta=2k\pi$$ is a pencil of five half-lines stemming from the origin.

Hence the roots are the five intersection points, forming a regular pentagon.

Answer 3

The argand plane is a plane, pretty much the same as the Euclidean plane except that $x$-axis is called the "real axis" and the $y$-axis is called the "imaginary axis". You can learn to visualize it, in your study of complex analysis.

You can also learn to visualize complex functions like $f(z) = z^5$. Rewrite this equation using polar form $z = r e^{\theta i}$, and you have $f(z) = r^5 e^{5 \theta i}$. Thus, the point $z$ in the plane is mapped to another point by taking the 5th power of its absolute value $r = |z|$ and by multiplying its angle $\theta$ by $5$.

Now, to solve the equation $z^5=1$, you can break it into two question:

First, which values of $r \in [0,\infty)$ satisfy $r^5=1$? Answer: only $r=1$.

Second, which values of $\theta \in [0,2\pi)$ satisfy $5\theta = 2 \pi n$ for some integer $n$? Answer: $\theta = 0$, $2 \pi / 5$, $4 \pi / 5$, $6 \pi / 5$, $8 \pi / 5$. So the solutions of $z^5-1=0$ are $$z = 1, e^{2 \pi i / 5}, e^{4 \pi i/5}, e^{6 \pi i/5}, e^{8 \pi i/5} $$