I have the following function where I want to identify the Riemann surface.
$$ f(z)=\log\left(\sqrt{z^2+1}\right). \quad\quad\quad (1) $$ The square root function has a Riemann surface $R_{SR}$ with branch points at $\pm i$ and two sheets. The log function has a Riemann sheet $R_{log}$ with a branch point at zero with many sheets. Since square root function maps $R_{SR}$ to $\mathbb{C}$
$$ \sqrt{\dots} : R_{SR} \rightarrow \mathbb{C} $$ The input for the log function in the equation (1) would be $\mathbb{C}$. However, this seems wrong since the log function should have $R_{log}$ as its domain.
So does this mean a composite function like equation (1) can not be smoothly defined?
Thank you
$$z\in \text{disk}(1,1), \quad z\to \log(1+z)$$ is analytic with derivative $$z \to \frac{1}{1+z}.$$ Inside this disk the algebraic formulae for positive exponents are valid $$\log \left((1+z)^n\right) = n \log \left( 1+z \right) $$ by differentiation, the chain rule for the derivatives and integration.
It follows that $$z \in \text{disk}(1,1), \quad z\to \log \left( (1+z^2)^ {\frac{1}{2}}\right)$$ is analytic with derivative $$z \to \frac{z}{1+z^2}.$$ The derivative has the two simple poles at $z\pm i$. The antiderivative $$z\to \frac{1}{2} \left(\log(1 + i z) + \log(1-i z) \right)$$ has two infinite log branch points at $z=\pm i$. As an integral of the sum of two single pole functions, it has a Riemann surface like the $\tan^{-1}$-function, but with different coefficients: a standard parkdeck connected by helix ramps winding around two towers.
$$\int \frac{dz}{1+z^2} = \tan^{-1}(z)= \frac{1}{2 i}\left(\log (1-i z)-\log(1+i z) \right)$$
Requisites to be used: Analytic continuation by the chain of overlapping circles, starting with circle of conveergency of the most simple Taylor or Laurent of any derivative or antiderivative.
Ambiguities of definitions by algebraic formula are discarded by covering the Riemann surface with circles of convergency.