Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
    
Is this a homework problem (for example, exercise 1 on pg 133 from Ahlfors)? If so, please use the homework tag. –  Adam Saltz Mar 13 '12 at 0:15
    
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
    
@IAmBrianDawkins Yes, this is that exercise from Ahlfors, but not homework. I've been working through the book on my own time. –  Dedede Mar 13 '12 at 0:19
show 2 more comments

3 Answers

up vote 3 down vote accepted

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.

share|improve this answer
add comment

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 $z=-\frac12$)!

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.