What shape does a quadratic trace if it is a function of a complex variable?

Question

I was thinking earlier, a quadratic like $y = x^2 + 3x + 2$ traces a parabola in the x-y plane. If we were to consider that x can also be a complex number, not just a real number, what shape does it trace in the argand diagram? I acknowledge that we do not have the ability to draw the graph since in this case it would need to exist in four dimensions, but can we understand the shape it would trace even though we can't strictly 'plot' it?

Answer 1

One interesting way to "plot" a quadratic like $f(x) = x^2 + 3x+2$ in the complex plane is to fix a radius $r$ and plot the complex points $f(r e^{i\theta})$ for $0 \le \theta < 2\pi$, which is a curve in the complex plane. We can do this for several values of $r$.

Here's a picture of these plots for $x^2 + 3x + 2$ in particular, when $r=0.1, 1, 2, 3, 4$. You should imagine an animation of this curve evolving over time, growing as $r$ goes from $0$ to $\infty$.

In purple, the $r=4$ curve shows you what the plot will look like for most large values of $r$: for a quadratic, they will be a curve that goes around the origin twice. (In general, a degree $d$ polynomial will go around the origin $d$ times, when $r$ is large.) The curve tends to "unwind" as $r$ becomes smaller: when $r$ is small, higher powers of $x$ contribute much less. The $r=0.1$ curve is tiny and blue in this picture: it's approximately a radius $0.3$ ($3$ times $0.1$) circle around the constant term of the quadratic.

We also see that when $r=1$ and $r=2$, the curve passes through the origin; this is what complex zeroes look like in this representation. (This corresponds to $x=-1$ and $x=-2$ being the roots.)

If such a plot of a degree-$d$ polynomial is a tiny circle around its constant term for small $r$, and winds around the origin $d$ times for large $r$, then intuitively it must pass through the origin $d$ times to get there, which is motivation for why all complex degree-$d$ polynomials have $d$ complex roots.