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How do I find the maximum value of $a$ such that $f(w) = w^2 + w$ is univalent in $|w|\lt a.$

I haven't a clue how to begin.


This is not a homework problem. It was assigned as a try problem.

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Have you thought about the points where $f'(w) = 0$? – Jesse Madnick Oct 24 '11 at 1:16
Yes Jesse; I have. I get $w=-\frac{1}{2}$. But I don't know what to do with it. – titi Oct 24 '11 at 1:51
up vote 2 down vote accepted

Suppose $f(r)=f(s)$. That's $$r^2+r=s^2+s$$ $$r^2+r+(1/4)=s^2+s+(1/4)$$ $$(r+(1/2))^2=(s+(1/2))^2$$ $$r+(1/2)=\pm(s+(1/2))$$ The plus gets us nowhere, the minus gets us $r+s=-1$, so as long as $r+s=-1$ has no solutions in $|w|\lt a$, you're OK. So, how big can $r$ and $s$ be (in modulus), and still avoid solutions of $r+s=-1$?

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Just wanted to add.. a geometric way to see this: for which disks centered at the origin does the map $z \rightarrow -1 - z$ take the disk to a disk disjoint from itself. – Zarrax Oct 24 '11 at 2:58
@Gerry I think you gave a complete solution to the problem. Since $r+s=-1$ we have that $|r|=|s+1|$ so as long as $|r|<\varepsilon$ and $\varepsilon<\frac{1}{2}$ the points $s$ satisfying your equation lives in $B(1,\varepsilon)$ which is disjoint from $B(0,\varepsilon)$... – Leandro Oct 24 '11 at 3:01
which indicates that $a=\frac{1}{2}$. Now for $\varepsilon>\frac{1}{2}$ you only have to produce one point satisfying your equation inside $B(0,\varepsilon)$. – Leandro Oct 24 '11 at 3:03

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