show that function is analytic

Question

Suppose that f is analytic in a domain $ C^{n-1} \times C $ and that for each $z' = (z_{1}, z_{2},...,z_{n-1}) $ $\in C^{n-1}$,the function $z_{n} \rightarrow f(z', z_{n}) $ has a unique zero $\alpha (z') $ in $C $. How to show that $\alpha $ is analytic. Any help please

Answer 1

Using the residue theorem, you can write $$\alpha(z') = \frac 1 {2\pi i}\oint_{|z_n| = r} dz_n \frac{z_n \partial f/\partial z_n (z',z_n)}{f(z',z_n)}$$ where $r$ is a suitably chosen radius. (You may have to work in a local region.)

The expression on the RHS is manifestly holomorphic in $z'$.