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.
Square Root of a Holomorphic Function
complex-analysis
Best Answer
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.