Largest disc around which this complex function is one-to-one? How would you determine explicitly the largest disc round the origin on which the function $f(z) = z^2 +z$ is one-to-one?
Is there a general method to do this for functions of this type? 
 A: 0) "Disk" shall mean disk centered at $0\in \mathbb C$.     
1) a) The equation $z^2+z=w^2+w$ is equivalent to $(z-w)(z+w+1)=0$.
So two different points $z,w\in \mathbb C $ will have the same image under $f$ iff $w=-z-1$.   
b)In every open disk $D$  of radius $\gt \frac {1}{2}$ there exists $\epsilon $ such that the points  $z=-\frac {1}{2}+\epsilon $ and $w=-\frac {1}{2}-\epsilon $  belong to $D$.
 Since $z$ and $w$ satisfy $z+w+1=0$ we have $f(z)=f(w)$,  so that $f$ is not injective on $D$.
Hence the required radius must be  $\leq \frac {1}{2}$ .     
c) On the other hand  if $z'$ is in the  disk centered at $0$ and of radius $ \frac {1}{2}$, then $z'-1$ is not in the disk.[Draw  a picture !].
Applying this to $z'=-z$, we conclude that  $f$ is is injective on that disk .     
d) Conclusion: the required radius is $\frac {1}{2}$
2) One could  make a change of variable $\zeta=z+\frac {1}{2}$,  and reduce the question to the radius of injectivity of $\zeta\mapsto \zeta ^2$ around $\zeta=\frac {1}{2}$.
This allows one to use that  the squaring function is injective on $\text {Re}  (z)\gt 0$.
3) More advanced users will notice that this example comes from a degree two covering of $\mathbb P^1$ ramified only at $z=\infty$ and $z=-\frac {1}{2}$ 
4) The injectivity of a holomorphic function on a disk is a very subtle problem.
Louis de Branges de Bourcia's solution of the Bieberbach conjecture on holomorphic functions injective on the unit disk is one of the most profound results in complex analysis proved in the twentieth century.
You can find a photo of the solver  sporting a fine Basque beret here.
A: You need to find the largest disk containing the origin for which there is no pair $(z,\omega)$ with $z^2+z = \omega^2 + \omega$
Notice you can rearange
$$z^2 + z = \omega^2 + \omega \Longleftrightarrow (z+\omega)(z-\omega) = (\omega-z)$$
If $z$ and $\omega$ are distinct we must have $z+\omega = -1$, so what can you say about the disk?
I don't think there's a way of doing it in general that will save any time, you just have to look at a few and learn some tricks. 
