What do polynomials look like in the complex plane?

Question

I have a hard time visualizing the fundamental theorem of algebra, which says that any polynomial has at least one zero, superficially I know this is true as every polynomial must have either an imaginary zero or real zero, but how do I visualize this in the complex plane?

For example if we have a real polynomial, we know that it is zero when it crosses the x axis this is because $y = 0$, however if $f(z) = 0$, then it must be the case that $f(z) = w = u+iv = 0+i0=0$ therefore every zero in $f(z)$ passes the origin? That does not make sense to me, what am I missing here?

Answer 1

If $f$ is a polynomial and $z\in\mathbb{C}$ then $f(z)$ also lives in $\mathbb{C}$ so we can not really plot $f$ because we need four dimensions (2 for the domain and 2 for the image).

But pick $z=a+b\,i$ fix $b$ and let's vary $a$ and plot the norm of $f(a+b\,i)$, which is a real number. Notice the if we change $b$ then we get another curve. Actually we have a continuum of these curves! The FTA says that at least one of this curves (and we have lots of them!) crosses zero.

Note: the norm in $\mathbb{C}$ is $f(z)=x+i\,y$ then $|f(z)|=\sqrt{x^2+y^2}$

Answer 2

The content of the fundamental theorem of algebra is that constant polynomials $p: \Bbb C \to \Bbb C$ are surjective. By the division algorithm we know that a polynomial of degree $n$ has $n$ roots counting multiplicity. Thus, I like to think of polynomials as covering maps. A polynomial $p$ defines a ramified $n$-fold cover of the complex plane by itself. That is, for each $c$, there are $n$ points $z$ such that $f(z)=c$, counted with multiplicity. The only time we have multiplicity is when $f(z)-c$ has repeated roots, which happens when $(f(z)-c)$ and $(f(z)-c)'=f'(z)$ share a root. Thus it happens at the roots of $f'(z)$, of which there are at most $n-1$. Thus polynomials define a ramified cover of $\Bbb C$ by itself ramified at a set of at most $n-1$ points.