Square Root of a Holomorphic Function Let $U$ be an open connected subset of $\mathbb{C}$ and $n\in \mathbb{N}$ be $even$. Let $z_1$, $z_2$,....$z_n$ be $n$ distinct complex numbers lying in the same connected component of $\mathbb{C}-U$. Show that there exists a holomorphic function $f$ on $U$ such that $(f(z))^2$=$(z-z_1)....(z-z_n)$ $\forall z \in U$.
I have managed to prove this result if $n=2$, but I am not able to genaralize it. Thanks for any help.
 A: As Daniel Fischer said, you can apply your result for $n=2$ and multiply the square roots of  $(z - z_{2k-1})(z - z_{2k})$, $k = 1, \ldots, \frac n2$.
Alternatively, you can show that if $z_1, \ldots, z_n$ lie in the same
component of $\Bbb C - U$ then there is a holomorphic function $g$
in $U$ such that
$$ \tag{*}
 g(z)^n = (z-z_1) \cdots (z-z_n) \, .
$$
If $n$ is even then the desired conclusion follows by choosing
$f = g^{n/2}$.
To show the existence of $g$, note that for any closed curve
$\gamma$ in $U$ the winding number $N(\gamma, z_i)$
is independent of $i$, so that
$$ 
 \frac 1n \int_\gamma \left( \frac{1}{z-z_1} + \ldots + \frac{1}{z-z_n}\right) \, dz = N(\gamma, z_1)
$$
is a multiple of $2 \pi i$. Therefore we can choose $z_0 \in U$ and define
$$
 g(z) = \exp \left( \frac 1n \int_{z_0}^z \left( \frac{1}{w-z_1} + \ldots + \frac{1}{w-z_n}\right) \, dw \right) \, .
$$
for any curve connecting $z_0$ with $z$
in $U$. The previous considerations show that the value is independent
of which curve is taken, so that $g$ is well-defined.
Then
$$
 h(z) = g(z)^{-n} (z-z_1)\cdots(z-z_n)
$$
satisfies
$$
\frac{h'(z)}{h(z)} = -n \frac{g'(z)}{g(z)} + \frac{1}{z-z_1} + \ldots + \frac{1}{z-z_n} = 0
$$
so that $h$ is constant in $U$. After multiplying $g$
with suitable constant, $(*)$ is satisfied.
A: Assume that $z_{2k-1}$ lies in the same connected component of $\Bbb C\setminus U$ as $z_{2k}$. Let $w_k(t)$ be a path that connects $w_k(0)=z_{2k-1}$ with $w_k(1)=z_{2k}$. Then the homotopy
$$
h(t,z)=\prod_{k=0}^{n/2}(z-z_{2k-1})(z-w_k(t))
$$
has at $t=0$ a square root
$$
g(0,z)=\prod_{k=0}^{n/2}(z-z_{2k-1})
$$
that has a unique continuation as square root $g(t,z)$ of $h(t,z)$. The function $g(1,z)$ solves the task.
