Showing that a function has a zero in Unit Ball centred at 0.

Question

Suppose that $f(z)$ is a non-constant entire function such that $$|f(z)|=1 \text{ for every } z \text{ with } |z|=1.$$

Then show that $f(z)$ does not have a zero in $\mathfrak{B}(0,1)$.

How I approached this one?

Define $$g(z)=\frac{1}{f(z)}$$ for $z \in \mathfrak{B}(0,1)$

As $$|f(z)|=1 \text{ for every } z \text{ with } |z|=1,$$ we also have that $$|g(z)|=1 \text{ for every } z \text{ with } |z|=1.$$

Since $f(z)$ and $g(z)$ are analytic, they are continuous on $\frak{B}(0,1)$. Thus by the Maximum Modulus Principle, $|f(z)|$ and $|g(z)|$ must attain their maximum value on the boundary $|z|=1$ only.

Thus $|f(z)|<1$ for $z \in \mathfrak{B}(0,1) \text{ and } |g(z)|<1$ for $z \in \mathfrak{B}(0,1)$

By definition, $$g(z)=\frac{1}{f(z)} \implies |g(z)|>1 \text{ for } z \in \mathfrak{B}(0,1) $$

Thus $|g(z)|=1$. How do I complete this proof? Is there a contradiction here?

Answer 1

The maximum is indeed attained at the boundary. We now have two possibilities:

1. $\bigl\lvert f(w)\bigr\rvert=1$ whenever $\lvert w\rvert<1$. But then, since $\lvert f\rvert$ is constant, it is a standard Complex Analysis statement that then $f$ itself is constant.
2. $\bigl\lvert f(w)\bigr\rvert<1$ for some $w$ such that $\lvert w\rvert<1$. Then $f$ attains its minimum at some $z_0$ with $\lvert z_0\rvert<1$. If $f(z_0)\neq0$, then $g$ attains its maximum at $z_0$ then.