An analytic continuation of the square root along the unit circle

Question

I want to find an analytic continuation of the square root along the unit circle but I am not sure whether I am doing it correctly.

Let $C_0$ be the open disk of radius $1$ around $1$, and let $f_0:C_0 \to \mathbb{C}$ be defined as $f_0(re^{i \varphi})=\sqrt{r} e^{i \frac{\varphi}{2}}$, where $\varphi \in (-\pi,\pi]$. Let $\gamma: [0,1] \to \mathbb{C}$ be the path given $\gamma(t)=e^{2 it \pi}$. Find an analytic continuation of $f_0$ along $\gamma$, i.e. a sequence $(C_k,f_k)_{k=0}^{n}$ of analytic continuations $f_k$ of $f_0$ such that the $C_k$ cover the image of $\gamma$. Show that $f_n(1)=-f_0(1)$.

I tried to do this as follows. Since $f_0$ is holomorphic on $C_0$ we have the power series expansion

$$ f_0(z)=\sum_{m=0}^{\infty} a^{(0)}_m (z-1)^m \tag{1} $$

where $a^{(0)}_m=\frac{1}{m!} \frac{\partial^m}{z^m} \sqrt{z} \big|_{z=1}$. I wanted to analyticaly continue this series by defining $C_1=\{z \in \mathbb{C} \ | \ |z-e^{i \frac{\pi}{4}}|<1 \}$ and considering the function

$$ f_1: C_1 \to \mathbb{C}, \ f_1(z)=\sum_{m=0}^{\infty} a^{(1)}_m (z-e^{i \frac{\pi}{4}})^m. $$

where $a^{(1)}_m=\frac{1}{m!} \frac{\partial^m}{z^m} \sqrt{z} \big|_{z=e^{i \frac{\pi}{4}}}$ and $arg(z) \in (-\frac{3 \pi}{4}, \frac{5 \pi}{4}]$. Let $z=re^{i\varphi} \in C_0 \cap C_1$. With $z_1=e^{i \frac{\pi}{4}}$ I have

$$ \sqrt{z}=\sqrt{z_1} \sqrt{\frac{z}{z_1}} =\sqrt{z_1} \sqrt{1+\frac{z}{z_1}-1} \underset{(1)}{=}\sqrt{z_1} \sum_{m=0}^{\infty} a^{(0)}_m (\frac{z}{z_1}-1)^m =\sum_{m=0}^{\infty} \frac{\sqrt{z_1}}{z^m_1} a^{(0)}_m (z-z_1)^m $$

Since the series representation of $f_1$ is unique and since $f_0(z)=\sqrt{z}$ the functions $f_0,f_1$ agree on $C_0 \cap C_1$. By iterating the steps above I can define disks $C_2, C_3,...C_8$ with centers $e^{i k\frac{\pi}{4}}$ and corresponding holomorphic functions $f_k$, $k=2,...,8$, each time requiring $arg(z) \in (-\pi+k \frac{\pi}{4},\pi+k \frac{\pi}{4}]$ for $z \in C_k$. Upon considering the power series expansion centered at $e^{i \frac{8 \pi}{2}}=e^{i 2 \pi}$ I should get $f_8(e^{i 2 \pi})=\sqrt{1} e^{i \pi}=-1=-\sqrt{1} e^{i \cdot 0}=-f_0(1)$. Am I on the right track here or is there an error?

Answer 1

This approach should work, but it's considerably more complicated than writing down such functions explicitly. For instance, take:

• $f_0 (re^{i\theta}) = \sqrt{r} e^{i\theta/2}$ on $\Omega_0 = \{re^{i\theta} \in \mathbb{C} \mid r> 0, \theta \in (-\pi/2, 3\pi/4)\}$,
• $f_1(re^{i\theta}) = \sqrt{r}e^{i\theta/2}$ on $\Omega_1 = \{ re^{i\theta} \in \mathbb{C} \mid r>0, \theta \in (\pi/2, 5\pi/4) \}$,
• $f_2(re^{i\theta}) = \sqrt{r}e^{i\theta/2} $ on $\Omega_2 = \{ re^{i\theta} \in \mathbb{C} \mid r>0, \theta \in (\pi, 9\pi/4)\}$.

These $f_i$ are analytic on their domains of definition, $f_i, f_{i+1}$ agree on the intersections of their domains, and $f_2(1) = e^{i\pi} = -1 = - f_0(1)$.