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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.

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$f(z)$ 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 $(t_0-\epsilon, t_0)$.

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