Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$ Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$
Hint: First show that $f(z)$ is a polynomial.
Clearly one can not use Cauchy Estimates to prove that $f(z)$ is a polynomial, the other way is to prove that $f(z)$ has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ?
Thank you !
 A: Here is another way to phrase Igor Rivin's and Sourav D's answers, which might make them seem more elementary and "discoverable''.
The basic facts one should know are:


*

*a non-zero entire function has only finitely many zeroes contained in the unit disk (since the zeroes are a discrete subset of $\mathbb C$);

*if $|a| < 1$, then $(z-a)/(1 - \overline{a} z)$ is a holomorphic function on the unit disk, taking the unit circle to itself, and with a zero at $a$.
Combining these, we find that for our given function $f$ as in the question,
there is a polynomial $g$ with the same zeroes as $f$ (and with the same multiplicities), which also satisfies $|g| = 1$ when $|z| = 1$.
The ratio $f/g$ then has no zeroes on the unit disk, and still takes the unit circle to itself.  As Igor Rivin indicates in his comment, an argument with logs
and maximum modulus shows that this ratio is constant, and thus that $f$ is a constant (of modulus $1$) times $g$.  (Or, as Sourav D notes, you can apply max. and min. modulus directly to $f/g$.)
Finally, the fact that $f$ is entire puts pretty severe constraints on the values of $a$ that can appear in $g$! 

So basically, in order to apply the hint, you have to construct the candidate polynomial, and you do this
by a consideration of the zeroes of $f$ in the unit disk.  
A: Since $|f|=1$ on the unit circle and $f$ is entire, if $f$ is non-vanishing inside the unit disk then by maximum and minimum modulus principle $f \equiv 1$ inside the unit disk. Hence $f$ has at least one zero in the unit disk. Also $f$ cannot have infinite set of zeros inside as there can be no accumulation point of zeros either inside or on the unit disk. Hence $f$ is bounded and has a finite set of zeros inside the unit disk.
Any such function has a Blaschke product representation :
$$f(z) = z^m g(z) \prod_{j=1}^{\infty} \dfrac{-a_j}{|a_j|}\dfrac{a_j-z}{1-\bar{a_j}z}$$
assuming $f$ has a zero order $m$ at the origin and $a_j$ are the other zeros, counting multiplicity. $g(z)$ is a non-vanishing bounded holomorphic function in the unit disk. But according to the problem $|g(z)|=1$ on the unit circle. Hence by previous argument $g(z) \equiv 1$. Also $f(z)$ will have poles at $z = 1/\bar{a}_j$, but $f$ is entire. Hence $f(z) = z^m$ for $m \in \mathbb{N}$. 
PS: When I took a graduate course in Complex Analysis, Blaschke product representation was given as a homework exercise. I believe it will be in the scope of the syllabus you are following as well. 
A: $f$ is analytic in the closed unit disk, being entire, so it can be written as a Blaschke product, from which the result follows immediately (that it can be so written is not a very hard fact, whose proof is sketched in the wikipedia article).
A: For a low-tech way to show $f$ has to be a polynomial:
The function
$$
z \mapsto \frac{1}{\ \overline{f(1/\,\overline{z}\,)}\ }
$$
is analytic on the puntured plane and agrees with $f$ on the unit circle, so we have
 $$
f(z) = \frac{1}{\ \overline{f(1/\,\overline{z}\,)}\ }\,.
$$
If $f$ were to have a zero at some $z \ne 0$, then by the symmetry $f$ would have to have a pole at the point $1/\overline{z}$. This is impossible since $f$ is entire. Therefore the only possibility for $f(z) = 0$ is at $z = 0$. Again using the symmetry, this means that $\lim_{w \to \infty}f(w) = \infty$. (If $f$ has no zeros then by using the symmetry you see that it's bounded and so constant.)


There's a pretty well-known theorem that says  $\lim_{w \to \infty}f(w) = \infty$ is enough to show that $f$ is a polynomial. The details are: 1) Show using compactness that $f$ can only have finitely many zeros $a_1, a_2, \ldots, a_n$; 2) Divide out all the zeros, leaving an entire function with no zeros; 3) invert, giving a function which is bounded everywhere by a polynomial; 4) By extended Liouville's theorem , it is a polynomial, but with no zeros, so is a constant. So $\frac{(z - a_1)(z-a_2)\cdots(z-a_n)}{f(z)}$ is a constant, thus $f(z)$ is a polynomial.


We now know $f$ is a polynomial.
To show that $f(z)$ has to be a monomial, note that we already showed that $f$ can have at most one distinct zero, in which case it has to be at $z = 0$. So $f(z) = a_n z^n$. But by the criteria on the circle $|a_n| = 1$.
Therefore $f(z) = a_n z^n$ with $|a_n| = 1$.
