Proving analyticity at a point by showing continuity

Question

To show $\sqrt{z}$ is analytic on $\mathbb C-\{z \le 0\}$, Ahlfors shows that $w=\sqrt{z}$ is continuous on that domain. I don't understand what he means by saying once the continuity is established the analyticity follows by derivation of the inverse function $z=w^2$. What theorem do we use?

Answer 1

Here is a detailed argument: let $g(z)=z^{2}$ and $f$ be the given function. Then $\frac {g(f(z+h))-g(f(z)} z=\frac {z+h-z} h =1$. This can be written as $\frac {g(f(z+h))-g(f(z)} {{f(z+h)-f(z)}} \frac {f(z+h)-f(z)} h =1$. The first factor tends to $g'(f(z)$ by definition of derivative and the fact that $f(z+h)-f(z) \to 0$. It follows that $\frac {f(z+h)-f(z)} h$ has a limit. [Note that the denominator $f(z+h)-f(z)$ does not vanish for $h$ near $0$ because $f(z+h)=f(z)$ implies $g(z+h)=g(z)]$ .

This argument is quite general. Note that $g'(f(z))\neq 0$ is also required and this is obviously true in our case.

Answer 2

Another approach would be to simply note that by restricting the domain of $f$ to $\mathbb C\smallsetminus \{z \le 0\}$, we have defined a (principal) branch cut for the complex logarithm, and now $\sqrt z:=e^{{1\over 2}Log z}$ is analytic, being the composition of two analytic functions.