Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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... – user53153 Jan 3 '13 at 0:33
@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. – Javier Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example? – Javier Jul 21 '13 at 4:59

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 $\mathbb{R}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.