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!



Can curves induced by analytic maps wiggle infinitely across a line?
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 $[0,1]$.


