TITLE: The Riemann Surface of Logarithmic Function
QUESTION [5 upvotes]: Who can tell me why the compactification of RS of Log is just Riemann Sphere, please?

REPLY [3 votes]: The Riemann surface is just a copy of $\mathbb C$. Compactifying it (by adding one point) gives therefore the usual sphere.
In general, if $f:\mathbb C\to U$ is an surjective holomorphic function with non-zero derivative with image an open subset of $\mathbb C$, the Riemann surface of the inverse function $f^{-1}:U\to\mathbb C$ (which is in general not-single valued, so I am abusing notation a bit here...) is just the graph of $f$, the set $\Gamma=\{(z,f(z)):z\in\mathbb C\}$. The restriction of the first projection $\mathbb C\times\mathbb C\to\mathbb C$ gives a holomorphic function $\Gamma\to\mathbb C$ which is a biholomorphism. Your question involves the case wher $f(z)=\exp(z)$.
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