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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 !

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Certainly the functions $z\mapsto z^n$ work. The problem then is whether there are others. – Michael Hardy Jan 4 '14 at 1:39
@bryanj You are right , but I think that is equivalent to showing that $\lim_{z \to \infty} f(z) = \infty$ i.e. $f(z)$ has a pole at $\infty$, (Because its abs value has to grow outside that disk because otherwise it has to be a constant ) – the8thone Jan 4 '14 at 1:47
also note… – benh Jan 4 '14 at 1:50
up vote 6 down vote accepted

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 $|a_n| = 1$.

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Thank you, You mean that $f$ is a constant ? I think I am suuposed to prove that $f=z^m$. – the8thone Jan 4 '14 at 2:54
I tought the symmetry principle was only true for Linear Fractional Transformations, am I wrong ? – the8thone Jan 4 '14 at 4:06
Yes: The "Reflection Principle" is more general than that. It's "Schwarz Reflection" across the boundary of a domain. The trick is: Notice that the function involving the conjugates is in fact analytic (things involving conjugating twice usually are). But when $|w| = 1$, then $1/w = \overline{w}$. The uniqueness theorem gives the rest. – bryanj Jan 4 '14 at 14:02
@bryanj : I've deleted my comments. – Michael Hardy Jan 7 '14 at 19:16
@MichaelHardy : Me too. – bryanj Jan 7 '14 at 19:19

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.

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$f$ 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).

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Thank you very much, But this is a question in a qualifying exam where Blaschke product has not been covered in the textbook, and one is supposed to be able to prove it by using "simpler" complex analysis – the8thone Jan 4 '14 at 1:52
@Roozbeh-unity there is nothing complicated about this argument. Either $f$ has no zeros in the disk, in which case the maximal modulus principle applied to $\log |f|$ shows that $f$ is constant, or it does have a zero $a,$ in which case $f/((z-a)/(1- \overline{a} z))$ has the same property of being $1$ on the circle, and has one fewer zero. – Igor Rivin Jan 4 '14 at 1:55

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.

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