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:

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

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

enter image description here

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

Example 2:

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

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

enter image description here

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

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.

  • $\begingroup$ If I may ask, why not to visualize $f$ as a displacement of one object in the plane to another in the plane too? $\endgroup$ – dmtri Aug 22 '18 at 10:20
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    $\begingroup$ 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. $\endgroup$ – Hans-Peter Stricker Aug 22 '18 at 10:24
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    $\begingroup$ Riemann's Surfaces may be relevant in visualising multidimensional complex functions. This video reference might help - youtube.com/playlist?list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF $\endgroup$ – paulplusx Aug 22 '18 at 10:24
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    $\begingroup$ 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). $\endgroup$ – Arthur Aug 22 '18 at 10:24
  • $\begingroup$ 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"? $\endgroup$ – 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:

A few theorems are really helpful to understand complex analysis. I'm recommending to read about :

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


Finally, I can only recommend you a book that I received for Christmas from my parents. It's a wonderful book : https://www.amazon.com/Visual-Complex-Analysis-Tristan-Needham/dp/0198534469

It is so well written and it's so pretty. In fact, complex analysis can be totally linked to analysis in R^2, you just have to do it properly and consciously.

Good luck !


Riemann's Surfaces may be relevant in visualizing multi-dimensional (complex) functions. There is an excellent video series by welch labs in their playlist of Imaginary Numbers are Real.

Even though they start with the history and visualization of Complex functions, they move on to show 4-D Complex functions on 3-D Plane (not exactly the function, more like it's shadow) quite beautifully with computer animations. Take a look at it, it might help.


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