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Suppose $f(z) = z + a_2 z^2 + \cdots$ with $f$ univalent on the unit disc, $f(0) = 0$, and $f'(0) = 1$. Define $g(z)^2 = f(z)/z$ and $h(z) = zg(z^2)$. Then $h$ is univalent on the disc, with $h(0) = 0$ and $h'(0) = 1$. Note that $h$ is odd, and $h(z)^2 = f(z^2)$ for each $z$ in the unit disc. How do you see that $h(z) = z + \frac{a_2}{2} z^2 + \cdots$?

EDIT: Added missing hypotheses.

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    $h(z) = \sum \frac{a_i}{i} z^i$ will not be odd unless $a_i = 0$ for all $i\equiv 0\mod 2.$2011-09-12
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    What is the motivation for this problem?2011-09-12
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    It is used in the special case $n=2$ for the Bieberbach conjecture. See http://www.math.sunysb.edu/~bishop/classes/math401.F09/Zorn.pdf2011-09-12

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