I just presented the two famous theorems of Picard (Small and Great) in a graduate course, but I have not managed to discover a good number of interesting applications.

List of applications (rather straight-forward though) found so far:

  1. If a meromorphic function on $\mathbb C$ misses three values, then it is constant.

  2. The equation $f^n+g^n=1$ has NO non-trivial meromorphic in $\mathbb C$ solutions if $n\ge 3$.

  3. If $f$ is entire an 1-1, then it is linear.

  4. If $f,g$ are entire and $g'=f(g)$, then $f$ is linear or $g$ is constant.

Could you provide any interesting applications of these theorem?

I have asked this question in Mathematics StackExchange, but I only received one response.

  • $\begingroup$ The MSE question and the answer are here: math.stackexchange.com/questions/1019221/… $\endgroup$ Nov 13, 2014 at 8:54
  • $\begingroup$ Yes, I know this since I asked this question as well, and I have mentioned it in the OP. $\endgroup$
    – smyrlis
    Nov 13, 2014 at 10:44
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    $\begingroup$ I know. I just wanted to give the link so that if someone wanted to see the answer(s) at MSE, they would find it easily. $\endgroup$ Nov 13, 2014 at 13:12
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    $\begingroup$ @MalikYounsi: This proof is not correct, because as I said in my previous remark the statement is false for $n=3$. The argument you gave only implies that $f$ and $g$ are constant unless they have common poles (see the Proposition here: books.google.hu/…). So I am still wondering what is the answer to my question. $\endgroup$
    – GH from MO
    Nov 14, 2014 at 5:15
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    $\begingroup$ @GHfromMO Ah, I didn't realize you were talking about meromorphic functions. Good point! The argument works for entire functions, but I don't know how to prove the result for meromorphic functions using Picard's theorem. $\endgroup$ Nov 14, 2014 at 11:54

2 Answers 2


One of the reasons Picard's theorems are so famous is that they have many generalizations, and these generalizations have more applications. Let me mention 2 of the most useful generalizations, both good for a graduate course:

  1. Montel's theorem. The set of meromorphic functions omitting 3 values is a normal family. (There is a short and neat derivation of this from Picard's theorem via the so-called Zalcman Lemma (also known as Brody's lemma). This derives Montel from the STSATEMENT of Picard's theorem, no matter how you prove Picard's theorem itself.). Montel theorem in turn has many applications, the most famous one is to iterations of holomorphic functions. The Chapter in Montel's own book on this gives some basic results of this theory. If you prefer a more modern source, Thurston, Combinatorics of rational maps, in the book MR1500163 Complex dynamics. Families and friends. Edited by Dierk Schleicher. A K Peters, Ltd., Wellesley, MA, 2009. The first part of this paper gives an excellent introduction to Montel's theorem and the subject of holomorphic dynamics.

  2. Generalization of your item 2 says: if $F(x,y)=0$ is an algebraic curve of genus $>1$ and $f,g$ are two meromorphic functions such that $F(f,g)=0$, then $f$ and $g$ are constant. This is due to Picard himself, but does not have a standard name like "the third Picard's theorem".

For example, the irreducible differential equation $F(w',w)=0$, where $F$ is a polynomial can have meromorphic solutions only when $F(x,y)=0$ is of genus $0$ or $1$, and all solutions are either rational of trigonometric or elliptic functions (Weierstrass theorem).

  • $\begingroup$ 2. follows from the Uniformization theorem + louville, right? The holomorphic universal cover is the disk, and so any such $(f,g)$ would need to factor through the disk, which would then be constant by Louville. $\endgroup$ Nov 13, 2014 at 15:10
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    $\begingroup$ Uniformization theorem is a much deeper result than Picard's theorems. Picard proved all this when Uniformization was not yet STATED, namely in 1884. Uniformization theorem if proved completely will occupy most of a 1-semester grad course. $\endgroup$ Nov 13, 2014 at 15:18
  • $\begingroup$ And all three theorems of Picard are easy consequences of the Uniformization theorem. $\endgroup$ Nov 13, 2014 at 15:20
  • $\begingroup$ Yes. I guess my question is "Is there a much easier proof of 'the third Picard theorem'?" I suppose there must be if Picard proved it! I need to track down that paper... $\endgroup$ Nov 13, 2014 at 15:25
  • $\begingroup$ You don't have to track down that paper, because there even easier proofs than Picard's own:-) I mean for the Little and the Great Picard's theorems, and even for the "third Picard's theorem". $\endgroup$ Nov 13, 2014 at 22:21

In general, holomorphic maps $f: \mathbb{C} \rightarrow \mathbb{C}$ have no fixed points; but using Picard theorem we can show that $f\circ f$ always have fixed point.

Theorem (Fixed-point theorem) Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be holomorphic. Then $f \circ f: \mathbb{C} \rightarrow \mathbb{C}$ always has a fixed point unless $f$ is a translation $z \mapsto z+b, b \neq 0 .$


Suppose $f \circ f$ has no fixed points. Then $f$ also has no fixed points, and it follows that $$ g(z):=\frac{f(f(z))-z}{f(z)-z} $$ is entire function. This function omits the values 0 and 1 ; hence, by Picard, there exists a $c \in \mathbb{C} \backslash\{0,1\}$ with $$ f(f(z))-z=c(f(z)-z), \quad z \in \mathbb{C} $$

Differentiation gives $f^{\prime}(z)\left[f^{\prime}(f(z))-c\right]=1-c$. Since $c \neq 1, f^{\prime}$ has no zeros and $f^{\prime}(f(z))$ is never equal to $c$. Thus $f^{\prime} \circ f$ omits the values 0 and $c \neq 0$ by Picard, $f^{\prime} \circ f$ is therefore constant. It follows that $f^{\prime}=$ constant, hence that $f(z)=a z+b$. Since $f$ has no fixed points, $a=1$ and $b \neq 0$


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