Here are three domain plots of a complex polynomial of degree $5$. The left picture is very zoomed out, and the right picture is more zoomed into the zeroes. (Pictures are taken from Elias Wegert's book "Visual Complex Functions".) The color indicates the argument of the function; the modulus is not featured.

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Say we focus our attention on one color, like yellow. Then we can see that the yellow lines coming in from infinity seem to "end" at the zeroes of the polynomial. My confusion is: I cannot justify why this should be the case in general. Why is it true that every yellow line (that is, a line of constant argument) coming in from infinity should terminate at a zero of the polynomial?

It makes sense to me that around a zero of order $n$, the polynomial should look like $z^n$ -- that is, there should be $n$ yellow lines emanating from that point. What is not clear to me, though, is why every yellow line coming in from infinity should terminate at a zero in particular. Why can't it terminate at any other point? Any suggestions / hints would be greatly appreciated. Thanks!


Not quite, there're continous lines (not only yellow, but also cyan and magenta) cutting through the right-most zeroes in the pictures.

Also, the cyan or magenta line comes to one side of other zeros but leaves out with other colours.

Yellow looks brighter than other colours. See more about luminous flux.

Note that Hue[1/6]=yellow, Hue[1/2]=cyan and Hue[5/6]=magenta, it depends on the values of arguments for field lines lying on (usually differ by $180^\circ$ before and after cutting).

See also how the electrostatic field lines passing through the poles here.

Considering $f(z)$ near zero $c$ with multiplicity $m$,

\begin{align} z &= c+re^{i\theta} \\ f(z) & \approx f(c)+\frac{f^{(m)}(c)}{m!}(re^{i\theta})^m \\ & = \frac{r^m}{m!} f^{(m)}(c) e^{im\theta} \\ \arg f(z) & \approx m\theta+\arg f^{(m)}(c) \end{align}

The last term governs how the colour wheel is rotated about $c$.


Around a nonzero point, the argument is virtually constant.

$$f(z+\epsilon e^{i\theta})\approx f(z)+\epsilon e^{i\theta} f'(z)$$

and when $\epsilon\ll |f(z)|$,

$$f(z+\epsilon e^{i\theta})\approx f(z).$$

The condition $\epsilon\ll |f(z)|$ cannot be met at a zero.


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