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I came across the following problem :

Let $\displaystyle f$ be an analytic function defined on $\Bbb D=\{z \in \Bbb C:|z|<1\}$ such that range of $f$ is contained in the set $\Bbb C \backslash (-\infty,0].$ Then how can I prove that

There exists an analytic function $g$ on $\Bbb D$ such that Re $g(z) \leq 0$ and $g(z)$ is a square root of $f(z)$ for each $z \in \Bbb D$?

Can someone point me in the right direction? Thanks in advance for your time.

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$\Bbb C\backslash(-\infty,0]$ is precisely where the principal branch of the logarithhm is holomorphic. This implies that $Log(f(z))$ is a well defined holomorphic function on $\Bbb D$..

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  • $\begingroup$ But how Log(f(z)) is square root of f?? $\endgroup$ – neelkanth Jun 2 '15 at 7:22
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Hint: Since $f$ has no zeros in $\Bbb D$, it follows that $f'/f$ is also analytic in $\Bbb D$. $\Bbb D$ is simply connected so we can find an analytic function $g$ so that $g' = f'/f$ (why?). Find the relationship between $g$ and $f$. With some adjustments to $g$, we can find the desired function.

Note: This method implies the existence of a holomorphic logarithm if you aren't already familiar with this fact.

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