Images of curves passing through zeros of a complex polynomials For real polynomials, it's well-known that for zeros of even multiplicity the graph of the function "bounces off" of the x-axis, while for zeros of odd multiplicity it passes through the x-axis. Is the same thing true for complex polynomials?
I've been told a curve passing through a zero of odd multiplicity is mapped to a curve passing through the origin while a curve passing through a zero of even multiplicity "bounces off" of the origin. Is this a well-known result? Is there a theorem for this? If I wish to use this fact in a proof do I need to justify it or is this common knowledge? I'm also interested in a proof of this fact, even in the case of real polynomials I don't think I've ever seen this be proven, it was just taught as a "trick" to graph polynomials.
 A: The "bouncing" of the graph of a polynomial of even multiplicity over the reals is a direct consequence of the fact that $(-x)^{2n} = x^{2n}$ for any natural $n$. For a given polynomial $p(x)$ and root $r$ of multiplicity $2n$, you can see this behavior more clearly by constructing a particular approximating polynomial $\hat{p}(x) = \hat{p}_r(x)$ obtained by evaluating $x=r$ in all but the factors of $(x-r)$.
In other words, write $p(x) = q(x) \, (x-r)^{2n}$ for some polynomial $q(x)$ such that $q(r) \neq 0$. Then, we define the approximation $\hat{p}(x) = q(r) \, (x-r)^{2n}$. Since
$$
\lim_{x \to r} \frac{p(x)}{\hat{p}(x)}
= \lim_{x \to r} \frac{q(x)}{q(r)}
= 1, 
$$
we can see that $\hat{p}(x)$ asymptotically approximates $p(x)$ as $x \to r$. But $\hat{p}(x)$ is much simpler to understand. It is just the power function horizontally shifted and vertically scaled (and possibly inverted).
Here's an example with many factors, in a neighborhood of $r=1$:
\begin{align} 
p(x) &= \tfrac{1}{50}(x+2)(x+1)^3 \color{green}{(x-1)^2}(x-2)(x-3) \\
\hat{p}(x) &= \tfrac{1}{50}(1+2)(1+1)^3 \color{green}{(x-1)^2}(1-2)(1-3) 
= \tfrac{24}{25} \color{green}{(x-1)^2}
\end{align}
And here's what the graphs look like near $r=1$.



How does this generalize to the case of a complex valued function? Over $\mathbb{C}$, it's still true that for even integer powers $(-z)^{2n} = z^{2n}$, polynomials still factor (even more so!), and we can make the same local approximation near the root. However, there are now more ways to be near the root $r \in \mathbb{C}$ than just to the left or to the right, and hence the even power identity applies to more pairs of points.
Explicitly, consider $z = r + \varepsilon\, e^{i\theta}$ for any $\theta \in [0, 2\pi)$. This is a complex number a distance $\varepsilon$ from $r$ in the direction of the angle $\theta$ in standard position, i.e. $|z - r| = \varepsilon$. This is a whole circle's worth of points, and any point of this form has an antipodal point
$$
z_- = r + \varepsilon\, e^{i(\theta + \pi)} = r - \varepsilon\, e^{i\theta}.
$$
In the approximating polynomial, these pairs of antipodal points evaluate to the same number: $\hat{p}(z_-) = \hat{p}(z)$, so a semicircular path in the domain produces a copy of all the outputs around the root, and as you finish the circular path around the root in the domain, you retrace the identical output values.
This is much more difficult to visualize as you would need $2 + 2 = 4$ real dimensions to graph a complex-valued function of a complex variable, but there are alternative techniques to visualize such functions. The method of domain coloring is particularly useful here. In the neighborhood of a root of even multiplicity $2n$ of a complex polynomial, the color (argument $\theta$) will cycle $2n$ times, which guarantees that antipodal points nearby will have approximately the same color. This is also true of their magnitudes. This is the complex version of the "bounce".
For example, here's the domain coloring image near a root of multiplicity $2$, and you can see the colors cycle twice.

