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Let $f(z)$ be a nonvanishing analytic function on a simply connected region $\Omega$. Then there is an analytic function $g(z)$ such that $e^{g(z)}=f(z)$. Is there any specific formula for $g(z)$?

(By specific formula I mean, for example, on the region $\mathbb{C}-\{x\le 0\}$ we know $$\log^{[k]}{z}=\log{|z|}+i\arg{z}+i2k\pi$$ where $k$ is an integer, and $\log^{[k]}{z}$ is holomorphic on $\mathbb{C}-\{x\le 0\}$.)

EDIT: Let me make my question clear. I know that we can use integral to define $\log{f}$. But that's not what I'm looking for. Let me take this example to explain what I want:

Let $f(z)=z^9$ and $\Omega$ the region $Re(z)>1$. Then there is a holomorphic function $g(z)$ on $\Omega$ such that $e^{g(z)}=f(z)=z^9$ and that $g(x)=9\log{x}$ for real $x>1$. In this case the formula I want is: $$g(z)=9\log|z|+9i\arg z$$ where $\arg z\in (−\pi,\pi)$.

fan
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  • Try the integral (from a fixed point $z_0$ to $z$ - path doesn't matter thanks to the simple connectedness) of $\frac {f'(w)}{f(w)} dw$. – Yoni Rozenshein Mar 11 '13 at 01:50

1 Answers1

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Hint: Try defining your function of $z$ as an integral from a certain function to a fixed point $z_0$ (the well-definedness [i.e. path independence] of which comes from simple connectedness).

Alex Youcis
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