# Visualizing complex functions $f: \mathbb{C} \rightarrow \mathbb{C}$

The graph of a complex function $f: \mathbb{C} \rightarrow \mathbb{C}$ is a 3-dimensional object in a 4-dimensional space and thus hard to visualize, even when it's a smooth 3-dimensional surface.

A natural way to visualize it is by two graphs of two functions $r: \mathbb{R}^2\rightarrow \mathbb{R}, i: \mathbb{R}^2\rightarrow \mathbb{R}$ with $r(z) = \text{Re}(f(z)), i(z) = \text{Im}(f(z))$. These functions define - under appropriate circumstances - two 2-dimensional objects $\color{red}{S^f_r}$ and $\color{green}{S^f_i}$ in a 3-dimensional space and can be visualized in $\mathbb{R}^3$, at least when they are smooth 2-dimensional surfaces.

In general these two surfaces may or may not

• intersect with each other at some points in $\mathbb{R}^3$ giving a 1-dimensional object $\color{blue}{C^f}= \color{red}{S^f_r} \cap \color{green}{S^f_i}$ which lives in $\mathbb{R}^3$
• intersect with the plane $\mathbb{R}^2$ giving two 1-dimensional objects $\color{red}{C^f_r} = \color{red}{S^f_r} \cap \mathbb{R}^2$ and $\color{green}{C^f_i} = \color{green}{S^f_i} \cap \mathbb{R}^2$ which live in $\mathbb{R}^2$

These 1-dimensional objects can be straight lines (or sets of straight lines), circles (or sets of circles), arbitrary open or closed curves (or sets of those).

One thing is obvious: There are $z_0 \in \mathbb{C}$ with $f(z) = 0$ (i.e. $f$ has roots) iff $\color{red}{C^f_r} \cap \color{green}{C^f_i} \neq \emptyset$. Knowing that each complex polynomial has roots (the fundamental theorem of algebra), we know that $\color{red}{C^P_r} \cap \color{green}{C^P_i} \neq \emptyset$ for all polynomials $P$. (We know even more: $\color{red}{C^P_r} \cap \color{green}{C^P_i}$ is a point set of size less or equal the degree of $P$.)

Example 1:

$P(z) = z^2 -1,\ r(x,y) = x^2 - y^2 - 1 ,\ i(x,y) = 2xy$

$\color{red}{C^P_r} = \{ (x,y)\ |\ x^2 - y^2 = 1 \}$, $\color{green}{C^P_i} = \{ (x,y)\ |\ x = 0 \vee y = 0 \}$

$P(z) = 0 \leftrightarrow z = (1,0) \vee z = (-1,0)$

Example 2:

$P(z) = z^2 + 1, r(x,y) = x^2 - y^2 + 1 , i(x,y) = 2xy$

$\color{red}{C^P_r} = \{ (x,y)\ |\ x^2 - y^2 = -1 \}$, $\color{green}{C^P_i} = \{ (x,y)\ |\ x = 0 \vee y = 0 \}$

$P(z) = 0 \leftrightarrow z = (0,1) \vee z = (0,-1)$

What I'd like to know:

1. What can - beyond the fundamental theorem - be said about the shapes and the positions of $\color{red}{C^P_r}$ and $\color{green}{C^P_i}$ for polynomials $P$ in general terms (depending on the degree of $P$)? May (or even does) the fundamental theorem result from the characterizations of $\color{red}{C^P_r}$ and $\color{green}{C^P_i}$?

2. How may holomorphic functions (= analytic complex functions) be characterized in terms of $\color{blue}{C^f}, \color{red}{C^f_r}, \color{green}{C^f_i}$?

• Like this: "For a holomorphic function $f$ the set $\color{blue}{C^f}$ must be non-empty and such-and-such"?
• Like that: "When a curve $C$ is such-and-such there is a holomorphic function $f$ with $C = \color{blue}{C^f}$"?
• Might a holomorphic function $f$ possibly be fixed by its $\color{blue}{C^f}$ alone or by the pair $(\color{red}{C^f_r}, \color{green}{C^f_i})$?

For the learned mathematician the answers to these question may seem obvious ("How can you ask?"), for me they are not, sorry.

• If I may ask, why not to visualize $f$ as a displacement of one object in the plane to another in the plane too? – dmtri Aug 22 '18 at 10:20
• That's another way of visualization that might be helpful, but what will we see? First of all: a dense set of arrows. So one may start with arrows starting from points $\mathbb{Z}^2$. This may yield some geometrical patterns or symmetries already. – Hans-Peter Stricker Aug 22 '18 at 10:24
• Riemann's Surfaces may be relevant in visualising multidimensional complex functions. This video reference might help - youtube.com/playlist?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF – paulplusx Aug 22 '18 at 10:24
• The most common way that I have seen to visualize complex functions (at least on wikipedia) is to color the complex plane such that each point has a color corresponding to the function value at that point. See, for instance, the Gamma function or the Riemann Zeta function (these also have the contour plot of $|f(z)|$ drawn in). – Arthur Aug 22 '18 at 10:24
• But the colors are hard to interpret, aren't they? Is the color out of a two-dimensional color space? If so, how could we tell the real and the imaginary part of a "color"? – Hans-Peter Stricker Aug 22 '18 at 10:26

It's going to be hard to explain everything with writing a few lines. Complex analysis is alas, known as a hermetic subject, wonderful but impossible to visualize. This is a myth! Anyone that has a good understanding of this notion sees the complex functions. But writing about it in a short mathstack answer is harsh. I recommend you to start your investigation alone, and then come back with exact questions:

Some of them deal with the notion of conformal applications. They are also really helpful to visualize what's happening. For instance, Moebius transformation :