Basic Complex Analysis: Is $0^{1/2}$ undefined? So I'm having a tiny crisis here. With real numbers, when faced with an expression like $0^{1/2}$, we simply have that $0^{1/2} = 0$. In fact, $0^x = 0$ $\forall x\ne0$ for real values of $x$.My textbook, however, defines complex exponentiation as: 
$$z^c = e^{c \log{z}}, z \in \mathbb{C}$$
Thus if we're in the complex system, we have that $0^{1/2} = e^{\frac{1}{2} \log(0)}$, which is undefined for all branches of the log function... This incosistency is bothering me. This came about as part of a question that was asking why the principal branch of $z^{1/2}$ doesn't have a Maclaurin series, which I thought was as straightforward as the fact that the derivative doesn't exist at $z = 0$. My textbook claims, however, that $z^{1/2}$ has a singularity at $z = 0$. I don't like leaving tiny things like this unresolved in my mind, so my question I guess is.. Is it indeed the case that $0^{1/2}$ is undefined?
 A: The original definition $z^{1/2} = e^{\frac12\log z}$ depends upon a well-defined branch of the logarithm function, and therefore is not itself defined at $z=0$. However, you can show that $\lim_{z\to0} e^{\frac12\log z}=0$ for any choice of logarithm. Therefore, you are welcome to define $0^{1/2}=0$, and the resulting function will be continuous at $z=0$ (and wherever the logarithm branch is defined); however, there is no way to define a continuous (much less analytic) function in a neighborhood of $0$ that matches the original definition of $z^{1/2}$.
Notice that this depended on the fact that $c=\frac12$ is real and positive. With the function $z^i = e^{i\log z}$, for example, setting $0^i=0$ is no longer a reasonable choice, as it does not result in the function being continuous.
A: For technical reasons, when talking about a holomorphic function $f$, we want an open domain of definition. There is no way to define $z^{1/2}$ as a holomorphic function on a neighborhood of $0$, and for that reason it's probably better to leave $z^{1/2}$ undefined at $z=0$.
On the other hand, if we are talking about possible square roots of the complex number $0$, then of course $0$ is the only option.
A: When $c$ is real and strictly positive, you have 
 $$|0^c|=\lim_{z\rightarrow 0} |e^{c\log z}| = \lim_{z\rightarrow 0} e^{{\rm Re \ } c \log  z}= \lim_{z\rightarrow 0} e^{c \log  |z|} = \lim_{R\rightarrow +\infty} e^{-R}=0$$
in whatever way you let $z$ go to zero and whatever your choice of branch of $\log$. So there is no ambiguity in that case. When $c$ is not real positive (e.g. negative) there are choices to be made.
