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Here is the problem: find all functions that are everywhere analytic, have a zero of order two in $z=0$, satisfy the condition $|f'(z)|\leq 6|z|$ and such that $f(i)=-2$. Any hint is welcomed.

1 Answers 1

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Here is a hint: consider $f'(z)/z$.

Since $f(z)$ has a zero of order two at $z=0$, the derivative $f'(z)$ is also holomorphic, and $f'(0)=0$. Thus, you may write $f'(z)$ as $z\cdot g(z)$, with $g(z)$ holomorphic. Then, the bound in the statement tells you that $|g(z)|$ is bounded.

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    Ok, if I understand correctly, $|g(z)|\leq 6$ and $g(z)$ everywhere analytic imply by Liouville that $g(z)=c$ for some constant $c$. Then, $f'(z)=cz$ or $f(z)=c/2\cdot z^2+b=c/2\cdot z^2$, since $f$ has a zero in $z=0$. Finally, since $f(i)=-2$, we have $c=4$. Is this correct?2011-12-09
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    @QuArK21343: that seems right to me!2011-12-09
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    I wanted to write "You got it" but that's apparently too short for a comment.2011-12-09