A surprising theorem in complex variables

Here’s a strange theorem I ran across last week. I’ll state the theorem then give some footnotes.

Suppose f(z) and g(z) are two functions meromorphic in the plane. Suppose also that there are five distinct numbers a1, …, a5 such that the solution sets {z : f(z) = ai} and {z : g(z) = ai} are equal. Then either f(z) and g(z) are equal everywhere or they are both constant.


A complex function of a complex variable is merom0rphic if it is differentiable except at isolated singularities. The theorem above applies to functions that are (complex) differentiable in the entire plane except at isolated poles.

The theorem is due to Rolf Nevanlinna. There’s a whole branch of complex analysis based on Nevanlinna’s work, but I’d not heard of it until last week. I have no idea why the theorem is true. It doesn’t seem that it should be true; the hypothesis seems far too weak for such a strong conclusion. But that’s par for the course in complex variables.

Update: I edited this post in response to the first comment below to make the theorem statement clearer.

8 thoughts on “A surprising theorem in complex variables

  1. That is a strange theorem. I’m a bit puzzled by the bit where “f(z) and g(z) are constant or they are equal everywhere”. This seems to suggest that when they are constant they are not equal everywhere. But surely they must be since they are both constant and equal on the points a_i?

    It is a surprising result though but, as you say, this sort of strangeness is often the case in complex analysis. The fact that the holomorphic functions are, thanks to the Cauchy-Riemann equations, are completely defined by their values on any small open set is also pretty weird.

    In some sense, this result of Nevanlinna’s seems to be a generalisation of that well-known property in that you need only five points rather than an open set and that it holds for meromorphic functions.

  2. Mark, the two functions could be constant without being equal. Say f(x) is constantly 5 and g(z) is constantly 6. Then f(z) = a and g(z) = a have the same solution set, namely the empty set, for a = 3 or any value of a other than 5 or 6.

  3. I hadn’t state the theorem very clearly. Thanks for helping me see that. I updated the post.

  4. And my objection (I just tried to write it, and saw my mistake) was mixing up inputs and outputs also. I was thinking about equal on 5 domain values, but this is saying pick 5 output values which the functions achieve for the same set of input values.


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