What is the Taylor series of $\sqrt{z}$ around $z=-1$

Question

How do I calculate the Taylor series of $\sqrt{z}$ around $z=-1$ in the branch that implies $\sqrt{-1}=+i$ ?

I didn't find an answer in the web.

Answer 1

Just like any Taylor series.

$$(\sqrt z)'=\frac1{2\sqrt z},$$ $$(z^{1/2})^{(n)}=\frac{(-1)^{n-1}(2n-3)!!}{2^n}z^{1/2-n},$$

Then

$$\sqrt{-1}=i,$$ $$\frac1{2\sqrt{-1}}=-\frac i2,$$ $$\cdots\\$$ $$((-1)^{1/2})^{(n)}=\frac{(-1)^{n-1}(2n-3)!!}{2^n}(-1)^{1/2-n}=-i\frac{(-1)^{n-1}(2n-3)!!}{2^n}.$$

For coherence, $(-1)^{1/2}=i$ everywhere.