Holomorphic square root 
Let $A_{R,r}=\{z\in\mathbb C: r\lt|z|<R\}$. Prove that there can not be a function $q \in O(A_{R,r})$ such that $q^2(z)=z$

$z=a\times e^{\phi i}, \; r\lt a \lt R, \phi \in [0,2\pi]$
and the square root would lool like this.
$q(z)=\sqrt{a}\times e^{\phi i/2}$
I don't understand why this can not be a square root for every $z \in A_{R,r}$.
 A: That function is no function at all, since it maps $a\left(=a\times e^{0\times i}=a\times e^{2\pi i}\right)$ to two different numbers, $\sqrt a\left(=\sqrt a\times e^{0\times i}\right)$ and $-\sqrt a\left(=\sqrt a\times e^{\pi i}\right)$.
Actually, there is no continuous square root function from $\{z\in\mathbb C\mid\lvert z\rvert=a\}$ into $\mathbb C$. In fact, suppose that $r$ is such a function and define$$\begin{array}{rccc}f\colon&\{z\in\mathbb C\mid\lvert z\rvert=a\}\setminus\{a\}&\longrightarrow&\mathbb C\\&ae^{i\theta}\ \bigl(\theta\in(0,2\pi)\bigr)&\mapsto&\sqrt ae^{i\theta/2}.\end{array}$$Then, for each $z\in\{z\in\mathbb C\mid\lvert z\rvert=a\}\setminus\{a\}$, $\left(\frac{r(z)}{f(z)}\right)^2=1$ and therefore $\frac{r(z)}{f(z)}=\pm1$. But, since $\frac rf$ is continuous and its domain is connected, it follows that you always have $\frac{r(z)}{f(z)}=1$ or you always have $\frac{r(z)}{f(z)}=-1$. In other words, either$$\bigl(\forall z\in\{z\in\mathbb C\mid\lvert z\rvert=a\}\setminus\{a\}\bigr):r(z)=f(z)$$or$$\bigl(\forall z\in\{z\in\mathbb C\mid\lvert z\rvert=a\}\setminus\{a\}\bigr):r(z)=-f(z).$$But this is impossible, since we are assuming that $r$ is continuous and therefore, in particular, it is continuous at $a$. And the limit $\lim_{z\to a}f(z)$ doesn't exist (we get $\sqrt a$ if we $z$ approaches $a$ through the upper halfplane and we get $-\sqrt a$ if $z$ approaches $a$ through the lower halfplane).
Since there is no continuous square root function from $\{z\in\mathbb C\mid\lvert z\rvert=a\}$ into $\mathbb C$, there is no continuous square root function from $A_{R,r}$ into $\mathbb C$ and therefore there is no holomorphic square root function from $A_{R,r}$ into $\mathbb C$.
