Power series of $\sqrt {z- \mathrm i}$ around $z=1$.

Question

I want to solve the following problem:

Expand $\sqrt {z+\mathrm i}$ as a power series of $z-1$ and give its radius of convergence. What is the biggest disk where the function and the series coincide?

I suppose that the function and the series coincide in a disk centered at $z=1$ with the radius of convergence of the power series as the radius of the disk.

Is there a quick way of obtaining the general expression or do I have to derivate, evaluate at $z=1$, repeat the process for each derivative and apply induction?

It confuses me that I don't see a way to put $\sqrt {z+\mathrm i}$ in terms of power series that I already know.

Answer 1

Let $z=w+1$. Then we want the power series in a neighbourhood of the origin of $$f(w) = \sqrt{(1+i)+w} = \sqrt{1+i}\sqrt{1+\frac{w}{1+i}}\tag{1} $$ and since $1+i = \sqrt{2} e^{i\pi/4}$ and $$ \sqrt{1+z}=\sum_{n\geq 0}\binom{\frac{1}{2}}{n}z^n = \sum_{n\geq 0}\binom{2n}{n}\frac{(-1)^{n+1}}{(2n-1)\,4^n}z^n \tag{2} $$ we have: $$ f(w) = 2^{1/4}e^{i\pi/8}\sum_{n\geq 0}\binom{2n}{n}\frac{(-1)^{n+1}}{(2n-1)4^n\sqrt{2}^n}e^{-i\pi n/8} w^n\tag{3} $$ and the radius of convergence is the distance from the closest (to the origin) singularity of $f(w)$, located at $w=-(1+i)$, from which $\rho=\color{red}{\sqrt{2}}$. That also follows from $$ \frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}.\tag{4}$$

Answer 2

$$\sqrt{z-i} = \sqrt{z\left(1 - \frac{i}{z}\right)} = \sqrt{z}\sqrt{1 + \frac{1}{iz}}$$

Then use the series expansion of $1 + X$ where $X = \frac{1}{iz}$

Answer 3

binomial expansion:

$(a+b)^k = a^k + k a^{k-1}b + \frac {k(k-1)}{2} a^{k-2} b^2 \cdots$

$((i+1)+(z-1))^{1/2} = ((i+1)^{\frac 12} + \frac 12 (i+1)^{-\frac 12} (z-1) \cdots$

$a_{n+1} = (\frac 12-n)(\frac 1n) (i+1)^{-1}a_n$