Power series square root for negative x

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 at the above site on page 2. $$\sqrt x = x\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2n)!}{4^{n}(2n-1)(n!)^2}\left(\frac{1}{x}-1\right)^{n},\ \Re (x)>1$$

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.

Solution 1:

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 $(-1)^{1/2}=i$.

Solution 2:

The exact series expansion of $\sqrt{x}$ for $x\geq1$ is :

$x-\sum_{k=1}^\infty \frac{(2k)!(x-1)^k}{(k!)^22^{2k}(2k-1)x^{k-1}}$

And the exact series expansion of $\sqrt{x}$ for $0\leq x\leq1$ is :

$1-\sum_{k=1}^\infty \frac{(2k)!(1-x)^k}{(k!)^22^{2k}(2k-1)}$

The explanations is included in my researches at:

https://www.researchgate.net/profile/Mones_Jaafar I hope you find it useful.