# How can the graph of a complex function be embedded in three dimensional space?

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.

• 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
• math.stackexchange.com/questions/301929/… – User3910 Aug 10 '13 at 14:30

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 $\mathbb{R}$.