Lets consider $f:D\rightarrow G$ a conformal map between two bounded domains $D,G$ and $\gamma$ a curve in $D$ with a finite length. Is it possible that $f \circ \gamma$ has an infinite length?
I guess no. Im thinking about something like the Peano curve which has infinite turns which could not work with $f$ since $f$ is angle-preserving. But this is just an idea.
On
The alternate case, the curve may approach the boundary.
$D$ is a region with boundary made up of three circular arcs, $G$ is a disk.
There is a conformal map from $D$ onto $G$ by the Riemann mapping theorem. Let it be chosen so that the cusp point in $D$ maps to the right-most point of $G$; let that point be $z=0$.
Consider the curve based on $y=x\sin(1/x)$ in $G$. Its length is $\infty$ based on comparison with $\sum 1/n$. Here it is shown in blue; the upper and lower bounding lines are $y=x, y=-x$.
In the corresponding curve $\gamma$ in $D$, the upper and lower bounding curves are tangent at the cusp.
The curve $\gamma$ has finite length, using limit comparison with $\sum 1/n^2$.
The answer is indeed “no” if the term “curve in $D$” is understood as a continuous function from a compact interval $[a, b]$ into $D$.
$f'$ is holomorphic (and in particular, continuous) in $D$. Therefore $|f'|$ is bounded on the (compact) image of $\gamma$ in $D$, say by a constant $M$. It follows that $\operatorname{length}(f \circ \gamma) \le M \cdot \operatorname{length}(\gamma)$ if the latter is finite.