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Suppose $f : \Omega \to \mathbb{C}$ is a holomorphic function on some open set $\Omega.$

If $f'(z)\neq 0$ for some $z\in \Omega,$ does there necessarily exist a neighborhood $U$ of $z$ where $f$ is injective?

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    Yes, and the inverse is holomorphic. This is a pretty standard theorem in complex analysis, the open mapping theorem. If you know multivariable analysis, just use the inverse function theorem and show that the inverse Jacobian satisfies the Cauchy-Riemann equations. – 2017-01-05
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    Using Cauchy-Riemann equations give a non-zero Jacobian, and the proposition follows from Inverse Function Theorem. – 2017-01-05

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Let's say $f'(a) = v \ne 0$ (I prefer to use $z$ for the variable rather than a specific point). Then $\text{Re}(\overline{v} f'(z)) > 0$ at $z=a$, and by continuity in some open disk $D$ containing $a$. Then I claim $f$ is one-to-one on $D$. In fact, this is true for any convex open set containing $a$ on which $\text{Re}(\overline{v} f'(z)) > 0$ for some $v$.

For any distinct $p_1, p_2 \in D$, $$f(p_2) - f(p_1) = \int_{C} f'(z)\; dz = (p_2 - p_1) \int_0^1 f'(p_1 + t (p_2-p_1))\; dt$$ where $C$ is the line segment from $p_1$ to $p_2$, and so $$ \text{Re}((\overline{p_2} - \overline{p_1})\overline{v} (f(p_2)-f(p_1)) = |p_2 - p_1|^2 \int_0^1 \text{Re}(\overline{v} f'(p_1 + t (p_2 - p_1)))\; dt> 0$$ so in particular $f(p_2) \ne f(p_1)$.