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My question is how to solve: Determine all entire functions $f$ with the property that if $|z|=1$, then $|f(z)|=1$.

I was thinking that I could solve this by first show that $f$ has to be a polynomial. Then I can use the formula \begin{align*} f(z)=z^{n}(a_{n}+a_{n-1}/z+...+a_{1}/z^{n-1+a_{0}/z^{n}}) \end{align*} where $a_{n}\neq 0$. And then somehow show that \begin{align*} f(z)=z^{n}a_{n}, \end{align*} with $|a_{n}|=1$. (I think that this is the answer.)

Could someone help me? What are the steps I should to?

Some theorems that my book covers and I think will be needed are Liouville's theorem, Maximum principle, fundamental theorem of algebra. Thanks!

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    Presumably we're restricting to functions that are analytic and, say, defined on some neighborhood of the unit disc in $\Bbb C$? – 2017-02-20
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    Please be more precise. You want to find holomorphic functions, not just any functions. Presumably these holomorphic functions are defined in some domain. We can try and guess at all of these things, but it would be better if you made them clear at the outset. – 2017-02-20
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    Sorry I forgot to write that $f$ should be an entire function! – 2017-02-22
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    You do realize that this makes the function map the unit circle to the unit circle (not necessarily onto?). – 2017-02-22

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Disclaimer: Wrong proof, see commments. I will leave it up for discussion purposes and edit it later. Think this is more about the maximum modulus principle. By that we have that $|f(z)|\leq 1$ within the unit disk. Also, similarly, by the minimum modulus principle, we have that the function either has a zero or we have $|f(z)|\geq1$.
So lets assume the function has no zero. Then we can conclude that $|f(z)|=1$ for all $z \in D$. Since $|f|$ is constant, so is $f$. So any function with $f(z)=w$, $|w|=1$ is a suitable candidate.
Lets assume the function has a zero of order n, lets say in $z_0$. Then we have $$ f(z)=(z-z_0)^nh(z) $$ Since $h(z)$ is zero-free within $D$, we can apply the min/max principle again. In absolute values, this will lead to: $$ \frac{1}{|z-z_0|^n} \leq|h(z)| \leq \frac{1}{|z-z_0|^n} $$
But that means that our function $f(z)$ statisfies the estimate for $z \in D$: $$ 1 \leq |f(z)|\leq 1 $$ As above, $f(z)$ has to be constant, but it has a zero. A contradiction.

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    I would preface by noting that your answer is highly dependent on holomorphicity (which is probably one of the only contexts this question would be asked in, anyway). – 2017-02-20
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    I assumed that the function is holomorphic, otherwise it would be a really long answer. Also, im currently double-checking my answer, as i somewhat think I did a mistake at some point. – 2017-02-20
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    Also, I assumed that the function is defined only on D and cont. up to the boundary. – 2017-02-20
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    Yeah I think that is a totally fair assumption. – 2017-02-20
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    What about $f(z) = z$? It satisfies $\lvert f(z) \rvert = 1$ iff $\lvert z \rvert = 1$ but is not constant. Am I missing something? – 2017-02-20
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    So is your conclusion that $f$ has to be always constant? Isn't a rotation not constant and still satisfies all the conditions? – 2017-02-20
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    Ill leave the answer up and edit it later. You are right. I will put a disclaimer at the beginning. – 2017-02-20
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Hint: For $a\in \mathbb D,$ the open unit disc, define

$$g_a(z) = \frac{a-z}{1-\bar a z}.$$

Then $g_a$ is holomorphic on a neighborhood of the closed unit disc, and $|g|=1$ on $\partial \mathbb D.$ Any finite product of such functions is a candidate here.

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    I don't understand how I should use this? Could you please explain abit further. – 2017-02-22