# image of complex continuous function cannot be a curve

Let $f : U \subset \mathbb C \to \mathbb C$ a continuous function where $U$ is an open and connected set.

Prove that the image of $f$ cannot be a curve on compex plane.

-
False without further conditions on $f$. Is $f$ analytic? Injective? Either would be sufficient. –  Erick Wong Jul 1 '12 at 20:46
@ErickWong: I found this exercise exactly as I post it. Can you give your solution with the further conditions ? –  passenger Jul 1 '12 at 20:50
math.stackexchange.com/a/44757/8157 (this requires analyticity of $f$) –  Giuseppe Negro Jul 1 '12 at 20:54

Are you sure there are no additional constraints, e.g. the Cauchy-Riemann equations are satisfied (continuity + C.R. -> analytic). One could simply take $f(z) = \Re(z)$, which is continuous but not analytic -- the image of an open set (say the ball of radius 1 around the origin) is a curve (the interval $[-1,1]$).
The reason this cannot hold for an analytic function is because of the open mapping theorem. Analytic functions are open mappings, so an open set $U$ has to map to an open set $f(U)$, which cannot be a curve.
Yes, $f(z) = \Re(z)$, as stated. The important thing to understand is that a curve is not open in the topology of $\mathbb{R}^2$. –  snarski Jul 1 '12 at 21:12