Thread: Maximum Modulus Theorem

Suppose that $\displaystyle f$ is entire and that $\displaystyle |f(z)-1|<1$ for all $\displaystyle z$ with $\displaystyle |z|=1$. Prove that $\displaystyle f(z)$ has no zeroes in the disk $\displaystyle \Delta= \{z:|z|<1\} $.

Originally Posted by tarheelborn

Suppose that $\displaystyle f$ is entire and that $\displaystyle |f(z)-1|<1$ for all $\displaystyle z$ with $\displaystyle |z|=1$. Prove that $\displaystyle f(z)$ has no zeroes in the disk $\displaystyle \Delta= \{z:|z|<1\} $.

Hint Denote $\displaystyle g(z)=f(z)-1$ and apply the Rouche's Theorem to $\displaystyle f(z)=g(z)+1$ .

We haven't had Rouche's Theorem. I have the Maximum Modulus Theorem and I am not really sure how to apply that here.

The maximum modlus theorem states if g is holomorphic, |g| reaches its maximum only on the boundary.
Now g=f-1 is holomorphic, and |g|<1 on |z|=1, so |g|<1 in the disk |z|<1.
So |f|=|1+g|>=|1-|g||>|1-1|=0 in the disk.