How can one find the largest disk about $0$ such that $z^2+z$ is injective? I'm trying to find the radius of the largest disk about the origin so that the map $f(z)=z^2+z$ is injective.
I know $f(0)=0$ and $f'(0)=1\neq 0$, so there is at least some disk of positive radius where $f(z)$ is injective. Also, $f(0)=f(-1)=0$, so the disk can have radius strictly smaller than $1$. 
I say if $f(z)=a$ is injective iff $f(z)-a=0$ has at most one root. The roots are 
$$
-\frac{1}{2}\pm\frac{\sqrt{1+4a}}{2}
$$
and thus are point on the opposite side of a circle in the complex plane centered at $-1/2$ with radius $|\sqrt{1+4a}/2|$. I'm stuck here. How can I explicitly find the largest radius of the disk around $0$ so that $f$ is injective on that disk? Thank you.
 A: I'm hoping I could add my own answer based on a few comments I received. Please let me know if my conclusions are misguided.
As lhf says, $f'(1/2)=0$, so $f$ is not injective around $1/2$, and thus the radius of the disk is strictly less than $1/2$. 
But suppose $z$ and $w$ are distinct points such that $|z|,|w|<1/2$. Then $|z+w|\leq |z|+|w|<1$. So $z+w\neq -1$ in particular. But then $f(z)\neq f(w)$, since otherwise, by Geoff Robinson's comment, $z+w=-1$, a contradiction. So $f$ is injective on $|z|<\frac{1}{2}$, and this is the largest such disk.
A: For $w=f(z)=z(z+1)$ to be injective, the inverse $z=f^{-1}(w)=-\frac12\pm\frac{\sqrt{1+4w}}{2}$ must be single-valued. This has an algebraic branch point at $w=-\frac14$ corresponding to $z=-\frac12$, so my guess would be that the radius is $\frac12$.
Another way of looking at this is that $z^2+z-w=0$ $\iff$ $1+4w=(2z+1)^2$. Now if $2z+1=re^{i\theta}$, then this equation is one-to-one for $\theta\in\left(-\frac\pi2,\frac\pi2\right]$ but no bigger interval, since then the square meets itself on the negative axis (centered at $-\frac12$). In other words, $f$ is in fact one-to-one on the half-plane $\{z\mid\Re z>-\frac12\}$. So the radius of injectivity about $0$ must be $\le\frac12$.
The only remaining question is whether $f(z)=f(-\frac12)=-\frac14$ has any other solutions. However, this is where the complex square root function, $z^\frac12$, comes to our rescue. It has a branch point at $0$, but only because the argument of nonzero values is not well-defined. For $0$, there is only one square root. Therefore, the radius is indeed $\frac12$, and $f$ is injective for $|z|\le\frac12$ (including the point at $z=-\frac12$)!
A: Expand $f(z)$ in a taylor polynomial around the points $z=-\frac{1}{2}$:
$$f(z)=-\frac{1}{4}+\left(z+\frac{1}{2} \right)^2 $$.
Suppose now that $f(z_1)=f(z_2)$; In that case we have
$$\left(z_1+\frac{1}{2} \right)^2=\left(z_2+\frac{1}{2} \right)^2. $$
Taking absolute values we see that $z_1,z_2$ have the same distance from $-\frac{1}{2}$.
If we substitue $z_1=-\frac{1}{2}+r e^{i \theta_1},z_2=-\frac{1}{2}+re^{i \theta_2}$, we find 
$$e^{2i \theta_1}=e^{2i \theta_2}. $$
Thus either the points $z_1,z_2$ coincide, or they lie on opposite rays from $-\frac{1}{2}$.
From this you can conclude that the largest such disk has radius $\frac{1}{2}$, as it's the largest one contained in a half plane about $-\frac{1}{2}$. 
