let $f$ be holomorphic on the unit sphere and $|f(z)| = 1$ for $|z| = 1$ and $f(-1) = 1$. Furthermore $f$ has no zero's, determine $f$

Question

let $f$ be holomorphic on the unit sphere and continous on the closure, suppose $|f(z)| = 1$ for $|z| = 1$ and $f(-1) = 1$. furthermore $f$ has no zero's, determine $f$.

So far i know with the maximum-modulus theorem that $f(x)$ has a maximum at the boundary which is 1. so $|f(z)| \leq 1$. I first used louisville's theorem, but $f$ is does not need to be holomorphic on $\mathbb{C}$. Is there any way of using the fact that $f$ has no zero's to determine $f$?

Mick

The unit sphere lives in three dimensions. Do you mean the unit disc? — Gregory Grant, May 25 '15 at 10:14
As is, $f$ is not uniquely determined. Are you sure the condition is not $\lvert f(z)\rvert = 1$ for $\lvert z\rvert = 1$? — Daniel Fischer, May 25 '15 at 10:16
If a holomorphic function has no zeros, which other holomorphic function does that suggest we might look at? — Daniel Fischer, May 25 '15 at 10:28
Essentially a duplicate of http://math.stackexchange.com/questions/127401/if-a-holomorphic-function-f-has-modulus-1-on-the-unit-circle-why-does-fz. — Martin R, May 25 '15 at 10:54

score 0 · Answer 1 · answered May 25 '15 at 10:34

0

Hint: if $f$ has no zeroes in the unit disc, then consider the function $\frac1f$ on the closure of the unit to deduce that $|f|$ takes it's maximum and minimum value, on the unit disc boundary. But $|f|$ is constant on the boundary, so...

answered May 25 '15 at 10:34

k1.M

5,577

score 0 · Answer 2 · answered May 25 '15 at 10:36

Let us define $g(z) =\frac{1}{f(z)} $ then $g$ is holomorphic and $|g(z)|\geqslant 1 $ into the unit disc and $|g(z)| =1$ on the bounduary hence by Maximum Modulus Principle $g(z) =\mbox{ constant }$ and since $g(-1) =1 $ thus $g(z)=1 $ for all $z.$

let $f$ be holomorphic on the unit sphere and $|f(z)| = 1$ for $|z| = 1$ and $f(-1) = 1$. Furthermore $f$ has no zero's, determine $f$

2 Answers2