Question about Taylor series expansion in complex numbers involving branch cut I want to consider a Taylor series expansion over complex numbers of $(i + z)^{-1/2}$ around $0$.
Using the usual formula I get
$$
(i + z)^{-1/2} = i^{-1/2} - \frac{1}{2} i^{-3/2} z + .... 
$$
I can see that the series converge around $|z| < 1$.
When I was computing this, I just used the formula. But then I realized $i^{-m/2}$ is apriori not well defined since there are more than one choice for this... so the above formula doesn't make sense as it is. How can I fix this issue? Choosing arbitrary choice of $i^{-m/2}$ for each $m$ doesn't sound a good idea.
 A: I would write
$$\begin{aligned}
(i + z)^{-1/2} &= \left(i(1 - iz)\right)^{-1/2}\\
&= e^{-i \frac{\pi}{4}}(1 - iz)^{-1/2}\\
&= e^{-i \frac{\pi}{4}}(1 + \frac{i}{2} z + \dots)
\end{aligned}$$
A: The function $\log(z+i)$ has a branch point at $z=-i$ and is multivalued on the plane.  We can cut the plane and choose a branch such that on chosen branch of the cut plane, $\log(z+i)$ is defined and analytic.

We shall cut the plane from the branch point at $-i$ along the negative imaginary axis to the point at infinity on the sphere.  And we will define $\log(z+i)$ such that $-\pi<\arg(z+i)\le\pi$.

Now, let $f(z)=(z+i)^{-1/2}\equiv e^{-\frac12\log(z+i)}$.
Note that on the chosen branch, $\arg(i)=\pi/2$ and we have $f(0)=e^{-\frac12 \log(i)}=e^{-i\pi/4}$.
Furthermore, $f$ is analytic on the cut plane with its Taylor series around $0$ and for $|z|<1$ given by
$$f(z)= \sum_{n=0}^\infty \frac{f^{(n)}(0)z^n}{n!}$$
Looking at the first few terms, we have
$$\begin{align}
f(0)&=e^{-i\pi/4}\\\\
f'(0)&=-\frac12 e^{-i3\pi/4}\\\\
f''(0)&=\frac34 e^{-i5\pi/4}
\end{align}$$
Can you finish now?
