The reason for the problem we get is that $0$ is the branch point of the square root function, and if we remove $0$, i.e. consider the map $f:\mathbb{C}^*\to \mathbb{C}^*$ sending $re^{i\theta}$ to $r^{1/2}e^{i\theta/2}$. It is a well-defined holomorphic map. We will prove that such $f$ cannot be extended to $\mathbb{C}$.
Proposition 1. There exists no square root function, which is continuous on $\mathbb{C}$.
Proof. Assume that such function $f$ exists, then actually $f$ is the inverse of the map $g:\mathbb{C}\to \mathbb{C}$ sending $z$ to $z^2$. Due to the inverse function theorem, $f'(z)\ne 0$ iff $z\ne 0$, and hence, for any $z\ne 0$, there exists $U_z, U_{z^2}$ are neighborhoods of $z$ and $z^2$, respectively, then $f|U_z=U_{z^2}$, and $U_z$ is homeomorphic with $U_{z^2}$, and the inverse map $f$ of $g$ is holomorphic in $\mathbb{C}^*$. If $f$ can be extended to $0$, by the assumption, $f$ is continuous at $0$, and now, by the Riemann's removable singularities theorem, we can extend $f$ to $\mathbb{C}$. But then $z=f(z)^2$, taking the derivative both sides at $z=0$, one gets $1=2f(0)f'(0)=0$, a contradiction. Hence, the square root function cannot be extended to the whole complex plane. (Q.E.D)
This reflects partly our desire, when we want to extend the domain of the holomorphic function to a larger domain. It is a very important process, for example, look at my previous note about the Riemann's zeta function, and see how its analytic continuation play a key role in proving the existence of infinite prime numbers, or computing some integrals. But in this note, we just focus on the simpler function, the square root.
Before doing this, we represent a common method to analytically continue a convergent power series $f_0$ on an open disk $D(z_0,r)$, where $D(z_0,r)$ is the open disk centered at $z_0$ with radius $r$. Pick any $z_1$ in this disk, we represent the series by changing the center, from $z_0$ to $z_1$. We then get a new series $f_1$, centered at $z_1$ with the radius of convergent is at least $r-|z_1-z_0|$ (draw a picture, then you can see this). But if we are lucky, the radius of convergence $r'$ of a new series $f_1$ is bigger than $r-|z_1-z_0|$, then we can analytic continue $f$ from a domain $D_0:=D(z_0,r)$ to a new domain $D_1:=D(z_1,r')$. And by doing that as far as possible, we obtain a chain $(f_0,D_0),...,(f_n, D_n)$, with $f_i|{D_i\cap D_j}=f_j|{D_j\cap D_i}$, i.e. they are the same on the intersection. One often do this on a path from $z_0$ to $z_n$, and $z_i$ are in this path.
But we still hope that there exists a function $f$ on $D:=\cup_{i=1}^n D_i$ such that $f$ is the analytic continuation of all $f_i$, i.e. $f$ is a holomorphic map, and $f|D_i=f_i$. Also, another question arises, if we analytically continue $(f_0, D_0)$ along a path $\gamma_0$ from $z_0$ to $z_n$ to obtain $(f_n, D_n)$, and then analytically continue $(f_0, D_0)$ along another path $\gamma_1$ from $z_0$ to $z_n$ to obtain $(f_n', D_n')$, then at a small neighborhood $U\subset D_n\cap D_n'$ of $z_n$, can we have $f_n|U=f'_n|U$? If this happens, by the famous identity theorem, one can conclude that $f_n$ and $f'_n$ is the same in $D_n\cap D_n'$. The answer of these questions are known under the name Monodromy Theorem.
Theorem 2. Let $G\subset \mathbb{C}$ be a region, $P,Q$ are two points of $G$. Let $f_0$ be a analytic function in $D_0:=D(P,r)\subset G$. Let $\gamma_0, \gamma_1$ is two path from $P$ to $Q$, such that $(f_0, D_0)$ can be analytically continued via $\gamma_0$ to $(f, D)$, where $D\subset G$ is an open neighborhood of $Q$, and $f$ is holomorphic in $D$. Also, $(f_0,D_0)$ can be analytically continued via $\gamma_1$ to $(D',f')$, where $D'\subset G$ is an open neighborhood of $Q$, and $f'$ is holomorphic in $D'$. Then $f|D\cap D'=f'|D\cap D'$ if $\gamma_0$ is homotopic with $\gamma_1$ in $G$. In the case $G$ is simply connected, and $(f_0,D_0)$ admits the analytic continuation along any path from $P$ to any point $D\in G$, then there exists a unique $F$ holomorphic in $G$ such that $F|D_0=f_0$.
As an example, we will do analytic continuation and then detect why the condition of simply connected domain is necessary via the square root function. For any $\alpha\in\mathbb{R}$, we can define
$${\alpha\choose k}:=\frac{\alpha(\alpha-1)...(\alpha-(k-1))}{k!}$$
And one can easily that this extends our familiar ${n\choose k}$, where $n$ is a natural number. Via this, we can define the binomial series for $z^{1/2}$ centered at $1$ as follow. One can see
$$f(z):=z^{1/2}=(z-1+1)^{1/2}=\sum_{k\ge 0}{1/2\choose k}(z-1)^k$$
And by the ratio test, this series has radius of convergence $1$. That means, for any $z_1\in D(1,1)$, where $D(1,1)$ is the open disk center at $1$ with radius 1, we have
$$f(z_1)=\sum_{k\ge 0}{1/2\choose k}(z_1-1)$$
Now, we want to represent the series with center $z_1$, and compute its radius of convergence. We first define if $z_1\in\mathbb{C}, z_1\ne 0$, and $z_1=re^{i\theta}$, then $\sqrt{z_1}:=re^{i\theta/2}$, i.e. $\sqrt{z_1}=f(z_1)$ . From this, if $z_1\in D(1,1)$, then for any point $z\in D(z_1, 1-|z_1-1|)\subset D(1,1)$, we have $\frac{z}{z_1}<1$, and we can represent the series $\sqrt(z)$ centered at $z_1$ as follows.
$$\sqrt{z}=\sqrt{z_1}\sqrt{\frac{z}{z_1}}=\sqrt{z_1}\sum_{k\ge 0}{1/2\choose k}(\frac{z}{z_1}-1)^k=\sqrt{z_1}\sum_{k\ge 0}{1/2\choose k}z_1^{-k}(z-z_1)^k$$
By ratio test again, this series has radius of convergence $1$. In particular, if we choose $z_1=e^{i\pi/4}$, which lies in $D(1,1)$, and lies on $S^1$, we have
$$\sqrt{e^{i\pi/4}}=e^{i\pi/8}\sum_{k\ge 0}{1/2\choose k}e^{-k\pi i/4}(z-e^{i\pi/4})^k$$
And because this series has radius of convergence 1, and $i\in D_1:=D(e^{\pi i/4},1)$ but $i\notin D(1,1)$, we now successfully analytic continue $f$ from the domain $D_0 := D(1,1)$ to the domain $D(e^{\pi i/4},1)$. Continue this process, with the point $i\in D(e^{\pi i/4},1)$ we obtain the new domain $D_2:=D(i,1)$. And respectively, our domain will be $D_3:=D(e^{3\pi i/4},1)$, $D_4:=D(-1,1)$, $D_5:=D(e^{5\pi i/4},1)$, $D_6:=D(e^{6\pi i/4, 1})$, $D_7:=D(e^{7\pi i/4},1)$. And we finally obtain again $D_8:=D(1,1)$ by doing analytic continuation on $D(e^{7\pi i/4},1)$. Let us denote $f_0:=f$, and for $i>0$, $f_i$ is the analytic continuation of $f_{i-1}$ to $D_i$. What we have just done is that we are doing analytic continuation along the circle centered at 0, with radius 1.
This finishes our example for doing analytic continuation by using power series along a path. Now, we turn to the failure of the Monodromy Theorem in this case. When we back to $(f_8, D_8)$ from $(f_0, D_0)$, we hope that $f_8(z) = f_0(z)$ for all $z\in D_0$. Unfortunately, by using the same series for computing $e^{i\pi/4}$, we have
$$f_8(z)=e^{\pi i/2}\sum_{k\ge 0}(e^{\pi i})^k{1/2\choose k}(z-e^{\pi i})^k=i\sum_{k\ge 0}(-1)^k{1/2\choose k}(z-(-1))^k=$$
$$=i\sum_{k\ge 0}{1/2\choose k}(-z-1)^k=i\sqrt{-z}=i\sqrt{-1}\sqrt{z}=-\sqrt{z}=-f_0(z)$$
So, what is wrong here? Actually, the reason is that $\mathbb{C}^*$ is not simply connected (can you prove that? As an exercise, you may wanna prove that $\pi_1(\mathbb{C}^*)\cong \mathbb{Z}$, i.e. the fundamental group is not trivial). Hence, the Monodromy Theorem does not hold in this case.
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