# 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.

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Hints: What can you conclude about $z$ and $w$ if $z \neq w$ but $z^2 + z = w^2 +w$? You may prefer to rearrange it as $z^2 -w^2 = (w-z)$. – Geoff Robinson Mar 12 '12 at 23:47
@GeoffRobinson Thanks for the hint. Dividing by $z-w$, I find $z+w=-1$. So $\mathrm{Im}(z)=-\mathrm{Im}(w)$ also. I'm not sure what to say after that. – Dedede Mar 12 '12 at 23:55
Certainly, the radius is less than $1/2$ because $f'(1/2)=0$ and $f$ cannot be locally injective around $1/2$. – lhf Mar 13 '12 at 0:18
Geometrically, $z+w=-1$ says the midpoint of the line segment joining $z$ and $w$ is $-1/2$. So what size of disk around $0$ will never contain both $z$ and $w$? – Robert Israel Mar 13 '12 at 0:21
@lhf (et al), I think f is injective on $\{z\mid\Re z>-\frac12\}\cup\{-\frac12\}$ so that the radius is exactly $\frac12$. Please see my argument below, and let me know if you disagree. – bgins Mar 13 '12 at 0:30

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.

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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 $z=-\frac12$)!

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Thanks bgins, I appreciate the extra details in this answer. – Dedede Mar 13 '12 at 0:32

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 $-\frac{1}{2}$.

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