I was searching for information about the power series representation for $\sqrt x$ valid for complex numbers so I found and 'hacked' this equation here: http://www.maeckes.nl/Reeksen/kwadraatwortel%20GB.html to make it work with complex numbers. It is now published on the above site on page 2.
$$\sqrt{x} = x\sum_{n=0}^{\infty}\frac{(-1)^{n-1}(2n)!}{(n!)^2(2n-1)}\left ( \frac{1}{4x}-\frac{1}{4} \right )^n,\ \ \ \ \Re(x)>1$$
Now I am looking for the corresponding equation for $\Re(x) < 0$ (or $\Re(x) < -1$), with two prerequisites:
1: No fractions in the exponents.
2: It must hold for complex $x$.
This is the closest I have been able to get so far:
$$\frac{1}{\sqrt{x+2}\sqrt{x-2}}=\sum_{n=1}^{\infty}\frac{(2 n-2)!}{((n-1)!)^2}(x+2)^{-n}$$
Any help is appreciated.
Best Answer
For $|-1/z-1|< 1$ ie. for $\Re(z) < -1/2$
$$i\sum_{n=0}^\infty {-1/2\choose n} (-1/z-1)^n=i(1-1/z-1)^{-1/2}= z^{1/2}$$ The RHS is the branch analytic for $\Re(z) <-1/2$ and such that $(-1)^{1/2}=i$.