Consider the Gauss hypergeometric function
$$_2F_1(a,b;c,z) = \sum_{n=0}^{+\infty} \frac{(a)_n(b)_n}{(c)_nn!}z^n, \quad |z| < 1, \quad (x)_n = x(x+1)\cdots(x+n-1), \quad (x)_0 = 1$$
The Encylopedia of Mathematics states without reference that the analytic continuation of $_2F_1(a,b;c,z)$ on $\mathbb C \setminus [1,+\infty)$ is univalent (i.e. injective), but I believe this is wrong. If $a$ or $b$ is a non-positive integer, then $_2F_1(a,b;c,z)$ is actually a polynomial, so multiple zeros are expected and it won't be univalent. And when $a,b \notin \{0,-1,-2,...\}$, Bieberbach's theorem implies that if $_2F_1(a,b;c,z)$ is univalent on $|z| < 1$, then $\Re(a+b-c-1) \leq 1$ because of the growth of the coefficients in the above power series. Maybe some hypotheses on the parameters $a,b,c$ are missing in The Encyclopedia of Mathematics article ?
In my case, I'm interested in knowing whether $_2F_1(a,b;c,z)$ is univalent in $\left|z - \frac{1}{2} \right| < \frac{1}{2}$ when $a,b \ll -c < 0 < c$. Has this been studied already ? Sufficient conditions for univalency would be proving either that $_2F_1\left(a,b;c,\frac{1}{2}z+\frac{1}{2}\right)$ is convex, starlike or almost convex on $|z| < 1$.
A toy model to study first could be $$_2F_1\left(a,a+\frac{1}{2};\frac{1}{2},z^2\right) = \frac{1}{2}\left[(1+z)^{-2a} + (1-z)^{-2a}\right]$$ where $a \ll -1$. As a first sanity check, one observes directly that zeros can only appear when $z$ lies on the imaginary axis (hence when $z^2 \leq 0$), which we exclude here.
EDIT : I thinks this criterion based on the Schwarzian derivative proves that the answer to my question is false.
