Constructing a function with N zeros inside the unit disk,

Question

Find all functions $f(z)$ such that:

a) $f$ is analytic in some region containing |z| $\le$ 1

b) $|f| = 1$ on $|z| =1$

c) $f$ has N simple zeros $z_1, ... , z_N$ inside $|z| < 1$ and no other zeros on $|z| \le 1$.

Any ideas are welcome.

I had initially thought that $z^N$ was a good candidate but it only fulfills parts a) and b); its zeros are at the origin, of multiplicity N.

Thanks,

Answer 1

There is a standard construction here called a Blaschke product that does this. Such products are constructed from the standard conformal map from $\mathbb{D} \to \mathbb{D}$ that swaps $0$ and $a$; that is,

$$f_a(z) = \frac{|a|}{a} \frac{a - z}{1 - \overline{a} z}$$

This map preserves the boundary of the disk, and hence has absolute value $1$ on the boundary of the disk (as do finite products of such maps).

Answer 2

Well, you still have to worry about finding $all$ such functions. To do this you will want to show that if $g$ is holomorphic on some $D(0,R)$ with $R>1,$ $|g|=1$ on $\{|z|=1\},$ and $g\ne 0$ on the closed unit disc, then $g$ is constant.