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I want to prove that the complex $Log(z)$ function is holomorphic, when not negative.

To prove it, I am asked to use a "helper" function, which I do not understand how it helps. The function is:

$$F(\omega)=\frac{w-w_0}{e^w-e^{w_0}}$$ When $w\neq w_0$. $$F(w)=e^{-w_0}$$

Otherwise. And in the function $w_0=Log(z_0)$.

How can I use this function to prove that the Log function is holomorphic. As previous knowledge there is a theorem saying that combinations of functions have a derivative $$D(f(g(z_0))=g'(f(z_0))f'(z_0)$$ Provided certain conditions are met. I also know that the derivative of $e$ is $e$.

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We have $\frac {Log\, (z_0+h)-Log\, z_0} h=\frac {Log\, (z_0+h)-Log\, z_0} {e^{Log\, (z_0+h)} -e^{Log\, z_0}}$. If you prove that $F$ is continuous, which is easy you see that above quantity tends to $e^{-Log \, z_0}=\frac 1 {z_0}$ which prove that $Log$ is differentiable at $z_0$ with derivative $\frac 1 {z_0}$. For this to work you have to make sure that Log is continuous at $z_0$ and for this you have to avoid $(-\infty,0]$.

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