Let $f:(t\_0-\varepsilon, t\_0+\varepsilon)\to\mathbb{C}$, be an analytic application, such that: $f(t)=0\Longleftrightarrow\ t=t\_0$.


Is it true that there is an analytic function $g:(t\_0-\varepsilon, t\_0+\varepsilon)\to\mathbb{C}$ with $g(t\_0)\neq 0$ so that: 


$\dfrac{f(t)}{|f(t)|}=g(t), \forall\ t\in (t\_0-\varepsilon, t\_0)$ and $\dfrac{f(t)}{|f(t)|}=\pm g(t), \forall\ t\in (t\_0, t\_0+\varepsilon)$?


I'm studying the bike tire tracks, and I want to show that the front wheel does not have singularities even if the rear one can have cusp singularities. I need a result of that type to prove the assumption.


I found it true for particular functions. Take $f(t)=(t, t^2)$, for example. Moreover, I have proved that the following two limits exists:


$\lim\limits\_{t\nearrow t\_0}\dfrac{f(t)}{|f(t)|}$ and $\lim\limits\_{t\searrow t\_0}\dfrac{f(t)}{|f(t)|}$ and they are either equal or opposite.



Interesting property of analytic functions
$f(z)$ is analytic in some disk around $t\_0$ in $\mathbb C$. Now 
$\overline{f(\overline{z})}$ is analytic, coincides with $\overline{f(z)}$ on the reals, and has the same order of zero as $f$ at $t\_0$. Thus 
$h(z) = f(z)/\overline{f(\overline{z})}$ has a removable singularity at $t\_0$ with a nonzero value there. Let $g(z)$ be one of the branches of $\sqrt{h(z)}$.
Since $(f/|f|)^2 = f/\overline{f}$, on $(t\_0 - \epsilon, t\_0)$ or $(t\_0, t\_0+\epsilon)$ we will have $f(t)/|f(t)| = g(t)$ or $-g(t)$ (and we can choose $g$ so that it is $g$ on $(t\_0-\epsilon, t\_0)$.


