Let $ \gamma:\left(a,b\right)\to\mathbb{C} $ be an analytic path (Meaning, near every $ t_{0}\in\left(a,b\right) $ the function $ \gamma $ has a Taylor expansion in $t-t_0 $ with positive radius of convergence).
Suppose that $ \gamma $ is injective and $ \gamma'(t) \neq 0 $ for all $ t $ .
Prove that $ \gamma $ extends to a holomorphic function is some complex neighbourhood $V \supset (a,b) $.
I'm not sure how to start. Any help would be highly appreciated. Thanks in advance.
For each $t_0\in(a,b)$, let $\sum_{n=0}^\infty a_n(t_0)(t-t_0)^n$ be the Taylor series of $\gamma$ centered at $t_0$ and let $r(t_0)\in(0,\infty)$ be such that that series converges to $\gamma(t)$ on $\bigl(t_0-r(t_0),t_0+r(t_0)\bigr)\cap(a,b)$. Let$$V=\bigcup_{t_0\in(a,b)}D_{r(t_0)}(t_0).$$Then $V$ is an open set and $V\supset(a,b)$. If, for any $z\in V$, you take $t_0\in(a,b)$ such that $z\in D_{r(t_0)}(t_0)$, and then you define$$\Gamma(z)=\sum_{n=0}^\infty a_n(t_0)(z-t_0)^n,$$you shall have extended $\gamma$ to an analytic function $\Gamma\colon V\longrightarrow\Bbb C$. Note that the definition of $\Gamma$ makes sense because, if $z\in D_{r(t_1)}(t_1)$ for some $t_1\in(a,b)$, then, by the identity theorem, $\Gamma(z_0)=\Gamma(z_1)$.