Suppose f is non-vanishing holomorphic function in $\mathbb D$ and if $|z|=1\to |f(z)|=1$ then we have to show that f is constant

I had learned to extend function using Schwartz reflection Priciple to lower half plane with the condition given
As per Hint Provided, I have to extend f to whole $\mathbb C$ by $f(z)=1/ \overline{{f(\bar{z})}}$ whenever $|z|>1$
I do not know How to proceed?
Any Help will be appreciated


My attempt:
[I am New to complex Analysis Just Learned Scwarz reflection principal,I had solved above using that .I wanted to check my approch. Have a look at solution ,If there is gaps in proof if you think , I will happy to see that]

As given $f$ is non vanishing continous function on closed Unit disc and holomorphic on unit disc.
We have to extend function all over c.
We have 3 partition , Inside disc , On boundary and outside disk
We had already definied function on first 2
for 3rd
$|z|>1$ take conjugate to make anticonformal $z\to \bar z$ thentake reciprocal $\bar z\to 1/\bar z$ now $|1/\bar z|<1$ so $f(1/\bar z)$ definied
To make it conformal map again we will take conjugate and reciprocal......[why reciprocal? see downward]
$1/\overline{f(1/\bar z)}$
$f:C\to C$ defined as
$$=1/\overline{f(1/\bar z)},|z|>1$$

Now time to use non vanishing assumption that means $f(z)\neq 0 for |z|<1$
by defination [use of reciprocal] for |z|>1 f(z) become bounded.
Now by symmetry principal we have extended funciton to whole $\mathbb C$ and we have bounded function.
By Lioveilie theorem it is constant


Since $f$ is holomorphic and non-vanishing on $\mathbb{D}$, $1/f$ is also holomorphic on $\mathbb{D}$. By Maximum Modulus Principle applied to $f$ and $1/f$, your condition implies that $$ 1 \leq |f(z)| \leq 1 \Rightarrow |f(z)| = 1 $$ for every $z\in \mathbb{D}$. Then $f$ achieves maximum modulus in the interior of its domain which implies that it is constant, again by Maximum Modulus Principle.

Another way to approach this is to use Schwarz Lemma, not the reflection principle.


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