If $g(z)$ is analytic function, and $g(z)=O(|z|)$ and g(z) is never zero then show that g(z) is constant.

Question

If $g(z)$ is analytic function, and $g(z)=O(|z|)$ and g(z) is never zero then show that g(z) is constant.

My attempt:

I could show using cauchy estimates that $g(z)=O(z) = a_0+a_1z$. Now g(z) is analytic and never zero it means $z \neq \frac{-b}{a}$. It means how can we conclude that it is constant?

Answer 1

Once you have that $g(z)$ is linear by the usual bounds for the coefficients of the MacLaurin expansion, the only way $g(z)$ has to avoid a zero is to be a non-zero constant.

By the fundamental Theorem of Algebra, this has a nice generalization: if $g(z)$ is holomorphic, non-vanishing and $O(|z|^m)$ for some $m\in\mathbb{N}^*$, then $g$ is a non-zero constant.