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?
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 = ["", ""])
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.
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 $$z = 1, e^{2 \pi i / 5}, e^{4 \pi i/5}, e^{6 \pi i/5}, e^{8 \pi i/5} $$