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Find the largest disk in which the mapping $f(z)=z^2+z$ is a one-to-one mapping.

If $f(z)$ is not one-to-one on a region $S$, we can find points $a\neq b\in S$ such that $a^2+a=b^2+b$. This means $a^2-b^2=b-a$, so that $a+b=-1$. Now, I think the question is asking for the largest disk centered at the origin; otherwise we can pick any disk inside the region $\{z:\Re z>2\}$.

So assume the disk is centered at the origin. If the radius is strictly greater than $1/2$, then clearly we can pick two distinct negative real numbers that sum to $-1$. So consider the disk $|z|\leq 1/2$. For any two points $a,b$ in the disk, we have $|a+b|\leq |a|+|b|\leq 1$. The only chance that $a+b=-1$ is if $a=b=-1/2$, but here we have $a=b$. So $|z|\leq 1/2$ is the largest possible disk.

Is this reasoning correct? I was first confused because of the ambiguous problem statement.

PJ Miller
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    Usually, one wants open disks, so your inequalities would then be strict. But your reasoning is correct, and good that you noticed that without fixed centre, there are arbitrarily large disks on which $f$ is injective. Alternative way to get the same result: $f(z) = \left(z+\frac12\right)^2-\frac14$. – Daniel Fischer Oct 18 '13 at 19:12

1 Answers1

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More generally, given a quadratic polynomial $p$ and a convex domain $\Omega$, the following are equivalent:

  1. $p'\ne 0$ in $\Omega$.
  2. $p$ is injective in $\Omega$.

Of course, 2 implies 1 for any holomorphic function in any domain. Conversely, suppose that 2 fails, i.e., $p(a)=p(b)$ for some $a,b\in\Omega$. Then $p-p(a)$ is a constant multiple of $(z-a)(z-b)$, which implies $p'((a+b)/2)=0$. Since $(a+b)/2\in \Omega$ by convexity, 1 fails.

user98130
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