Htykuut

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.

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 $mathbb{R}$.