Where is $\sqrt{e^z+1}$, $z \in \mathbb{C}$, analytic? I'm considering the following question:
Where is $f(z)=\sqrt{e^z+1}$, $z \in \mathbb{C}$, analytic? Find $f'(z)$ where it is analytic.
My approach has been to simply differentiate $f(z)$ to get $$f'(z)=\frac{\mathrm{e}^z}{2\sqrt{\mathrm{e}^{z}+1}}$$ and note that then $\mathrm{e}^z \neq -1$, giving $z \neq i(2k+1)\pi$, $k \in \mathbb{Z}$. But I realize that the square root isn't continuous on all of $\mathbb{C}$, so are there any intricacies that I should be paying attention to?
Looking forward to your replies!
 A: The term analytic often allows the function to be multivalued (i.e., to be a function of the path of analytic continuation). If this is the meaning of analytic in your post, then the answer is: "$f$ is analytic at every point except odd multiples of $\pi i$, and its derivative is given by the formula you found, where the branch of the square root  is the same as the one used for $f$."
If you insist on a single-valued function (which I would call holomorphic, to disambiguate from the above), then one needs some branch cuts. There is no unique "canonical" choice of branch cuts. One just needs to cut every closed path for which the sum of winding numbers   around   odd multiples of $\pi$ is an odd number.  
One way to do this is to remove line segments of length $2\pi$ connecting two consecutive multiples of the above form. That is, let
$$\Omega = \mathbb C\setminus \bigcup_{n\in\mathbb Z} [(4n-1)\pi i, (4n+1)\pi i]$$
where the notation $[a,b]$ means the line segment from $a$ to $b$ in the complex plane. The cuts ensure that every closed curve in $\Omega$ that winds around one odd multiple of $\pi i$  also winds around another one, with the same index. As a result, $e^z+1$ travels an even number of time around the origin, which makes its square root single-valued.
Another option, perhaps a simpler one, is to cut the plane along horizontal half-lines starting at every odd multiple of $\pi i$. The remaining domain is simply-connected, so the monodromy theorem applies. 
