5
$\begingroup$

In his book Visual Complex Analysis (an awesome book, by the way), Needham, on the topic of graphing complex functions, says that

Actually, the situation is not quite as hopeless as it seems. First, note that although two-dimensional space is needed to draw the graph of a real function $f$, the graph itself is only a one-dimensional curve, meaning that only one real number (namely $x$) is needed to identify each point within it. Likewise, altough four-dimensional space is needed to draw the set of points with coordinates $(x,y,u,v)= (z, f(z))$, the graph itself is two-dimensional, meaning that only two real numbers (namely $x$ and $y$) are needed to identify each point within it. Thus, intrinsically, the graph of a complex function is merely a two-dimensional surface, and it is this susceptible to visualization in ordinary three-dimensional space.

Is this actually possible? I've been thinking for a while about how one would graph a complex function using only three dimensions, but I can't find a way.

$\endgroup$
4
  • 1
    $\begingroup$ It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization... $\endgroup$ Commented Jan 3, 2013 at 0:33
  • $\begingroup$ @PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds. $\endgroup$ Commented Jan 3, 2013 at 0:59
  • $\begingroup$ @JoeHobbit: What do you mean? Could you give an example? $\endgroup$ Commented Jul 21, 2013 at 4:59
  • $\begingroup$ math.stackexchange.com/questions/301929/… $\endgroup$ Commented Aug 10, 2013 at 14:30

2 Answers 2

4
$\begingroup$

I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.

However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $\mathbb{R}$.

$\endgroup$
0
$\begingroup$

Notwithstanding the answers emphasizing how to interpret the explanations on Needham's book, perhaps it would be a good idea to see how the current version of Wolfram Mathematica deals with the issue: A function $f(z)$ may be visualized using the command ComplexPlot3D, https://reference.wolfram.com/language/ref/ComplexPlot3D.html The way it is done is given a complex domain $$z=x+yi;\,x, y\in \{a\leq x \leq b,\, c\leq y \leq d\},$$ $$a,\,b,\,c,\,d\in \mathcal{R}$$ Then each point in the graph is (z,|f(z)|), with $f(z)=u+vi$ in the codomain of $f(z)$ and the magnitude or modulus of the function defined as $$|f(z)|=\sqrt{u^2+v^2}$$ with a color function that depends on the argument of the function, that is $$\arg(z)=\arg(x+yi)=\varphi$$ where $$\varphi = \arg (x+yi) = \begin{cases} 2 \arctan\left(\dfrac{y}{x + \sqrt{x^2 + y^2}}\right) &\text{if } y \neq 0 \text{ or } x > 0, \\ \pi &\text{if } x < 0 \text{ and } y = 0, \\ \text{undefined} &\text{if } x = 0 \text{ and } y = 0. \end{cases}$$ The aforementioned color function is what takes care of the "fourth dimension", so to speak, of the complex function, and it is very insightful, because of the color pattern that indicates how does it look like the function around zeroes, poles and essential singularities, or as it is explained there, in the wolfram reference page, "The color function goes from $-\pi$ to $\pi$ counterclockwise around zeros, clockwise around poles and infinite cycles near essential singularities." (I recommend visiting the page and looking at the pictures of the color function). Summarizing and recapitulating: the solution for obtaining the graph of a complex function in Mathematica, is by means of plotting $(x+y i,|f(z)|)$ with a color function that depends on the argument $\varphi$. Notice the mixture of using Cartesian coordinates in the domain, with polar coordinates for the range in the co-domain, using only the magnitude or modulus of the function as a third coordinate, and essentially "reinterpreting" the fourth coordinate as a color pattern.

Notice that when you use the magnitude of the function as a third coordinate, usually what you take is the principal branch of the magnitude, i.e., the positive value of the square root. Alternatives to Mathematica can be found, for example, in Python, which also uses $|f(z)|$ for the third dimension:

3D visualization of complex functions with matplotlib

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.