Can curves induced by analytic maps wiggle infinitely across a line? Let $f$ be a function analytic on an open subset $D\subset \mathbb{C}$, and let $\gamma:[0,1] \to D$ be a line segment. $g = f\circ\gamma$ is another curve in the complex plane; is it possible to for $g$ to cross a straight line infinitely often, where the crossing points accumulate towards a point? That is, does there exist a point $\alpha$ and a ray $R$ emanating from the point $\alpha$ such that for all $\epsilon > 0$, $g$ crosses the ray $R$ infinitely many times in the $\epsilon$-ball around $\alpha$?
We're trying to show something about analytic continuation, but we cannot rule out pathological beasts like these.
Thanks!
 A: The image of $[0,1]$ is compact and so must contain the purported
accumulation point. It makes no loss to assume that $\gamma(t)=t$,
$f(0)=0$ is the accumulation point, and the line in question
is the real axis. Then $f(z)=a_n z^n+a_{n+1}z^{n+1}+\cdots$ where
$a_n$ is nonzero and $n$ is a positive integer.
At this stage I'll assume there is a sequence of reals $t_1>t_2>\cdots$
tending to zero with each $f(t_j)$ real. we want to show that all the $a_k$
are real. Then considering $f(t_j)/t_j^n$ we get $a_n$ real. Now consider
$f(z)-a_n z^n$ in place of $f(z)$. We get $a_{n+1}$ real etc. So $f$
takes reals to reals so there's no "crossing" of the real axis.
In general, there must be a sequence of distinct points $(t_j)$ in $[0,1]$
where the curve crosses the line and which tends (by Bolzano-Weierstrass)
to a point $t\in[0,1]$. Replace $[0,1]$ by $[0,t]$ or $[t,1]$
(one of these has infinitely many $t_j$) and rescale the interval
to $[0,1]$.
A: I am little worried by Gowers' comment as this seems fairly straightforward to me.
Let $L:{\mathbb C} \to {\mathbb R}$ be a nonconstant affine map vanishing on your ray.  I assume that by "line segment" you really mean that $\gamma$ is linear (or affine).  In any case, as long as $\gamma$ is real analytic (by which I mean that it extends to a real analytic function on some open neighborhood of $[0,1]$), so is $h := L \circ f \circ \gamma$.  A real analytic function on a compact interval is either always zero or has only finitely many zeros.  If $h$ is identically zero, then your curve will stay on the line containing your ray and can enter or leave the ray only finitely many times.  If $h$ is not identically zero, then your curve meets the ray only finitely many times.
