Geometry of images of maps $f: \mathbb R \to \mathbb C$? I am having trouble seeing what a continuous map $f: \mathbb R \to \mathbb C$ might look like. If it was linear it would look like a line but it's not clear to me what happens if it's any map. 
I started to wonder about this after I was trying to work out what a convolution of two such maps might look like.
Then I ended up thinking about characteristic functions. These of course do look like just a line of the length of the set of which they are the characteristic function of. But are there maps for which the image is not a line but a plane?

What can images of continuous maps $f: \mathbb R \to \mathbb C$ look
  like?

 A: The Hahn-Mazurkiewicz theorem says (among other things) that a subset of $\mathbb{C}$ is a continuous image of $[0,1]$ iff it is compact, connected, and locally connected.  This includes, for instance, any closed disk, as well as many other more complicated sets that don't look anything like a "curve".  If you consider continuous images of $\mathbb{R}$ rather than $[0,1]$, you get exactly those sets which are path-connected and a union of countably many compact, connected, and locally connected sets (see this answer on MO, for instance).  In particular, for instance, $\mathbb{C}$ is the continuous image of some function $\mathbb{R}\to\mathbb{C}$.
Probably the best-known explicit construction of a curve in the plane whose image has nonempty interior is the Peano curve which is described nicely on Wikipedia.  The key technical tool in its construction is the fact that a uniform limit of continuous functions is continuous; the curve is constructed as the limit of a sequence of polygonal curves that get increasingly "dense" in a square.
Note that all of this is about continuous maps.  Given the tags you chose for the question, it sounds like you might actually be more interested in differentiable maps.  In that case, the story is quite different.  In particular, for instance, if $f:\mathbb{R}\to\mathbb{C}$ is $C^1$ (or more generally, locally Lipschitz), then it is not hard to show the image of $f$ must have Hausdorff dimension $\leq 1$.  In particular, this means it must have measure zero.
A: Every continuous map $ f: \mathbb R \to \mathbb C $ is a curve in complex plain, in particular a map $f:[a,b] \to \mathbb C$ " $a$ and $b$ are real numbers", is called a path,  which is used in complex integration theory. But about your second question I have to say that no! No such a map has a plain as its image. A simple example of a such map is $$f: [0,2\pi] \to \mathbb C$$                      $$f(t)=e^{it}$$ which its image is a circle.
