Positive real part of derivative imply injectivity? Suppose that $U\subset\mathbb{C}$ is open, connected, and that the derivative of the analytic functions $f:U\to\mathbb{C}$ has strictly positive real part.  Does it follow that $f$ is injective on $U$?
I am trying to just prove this directly by taking a path in $U$ from $x$ to $y$ (where $f(x)=f(y)$, and showing that $\Re f'>0$ implies $x=y$, but cannot seem to find a way.
If it is not true in general, how about bounded away from zero, that is, if the condition is $\Re f'>1$ on $U$?
 A: The theorem of Noshiro-Warschavski gives the result if $U$ is convex, while if $U$ is simply connected and  not convex, there are always counterexamples as per the Proceedings of AMS paper of Herzog and Piranian (pdf link), with the first page already giving the simple $f(z)=z^{1+\frac{\pi}{2\beta}}, f(1)=1$ for the domain $-\beta < \arg z < \beta$ where $\beta$ is fixed s.t. $\frac{\pi}{2} < \beta \le \pi$
The Noshiro-Warschavski follows immediately by integrating $f'$ on the straight line between any two points $x,y$ in $U$ which is allowed since $U$ convex (if more details needed, I can supply them, but it is a very simple exercise and I recommend trying to do it).
Note that if given non-convex open connected $U$ and an analytic function $f$ on $U$ with $Re f' >0$ and two points $f(z_1)=f(z_2)$, we can enclose them in an open subset $V$ with compact closure in $U$, so $\Re f' \ge a >0$ on $V$ and by scaling we can make $Re f'$ as high as we want on $V$ and still have $f(z_1)=f(z_2)$, so no bounding away condition can work if the domain is not convex
