The following is part of an exercise that I don't quite get:

Let $\phi: \mathbb D\setminus \{0\} \to \mathbb C$ be an injecitve holomorphic function with Laurent series $$\phi(z) = \frac1z + \sum_{n=1}^\infty a_nz^n$$ Prove that $$\sum_{n=1}^\infty n|a_n| \le 1$$

Hint: Calculate the area of $S_r = \phi(B_1(0)\setminus B_r(0))$ using Stokes' thoerem and consider the limit as $r \to 1$.

I have to say, I'm a bit clueless here. I don't even really see how to calculate the area of $S_r$. It seems that Stokes' theorem forces me to consider the real and imaginary parts of $\phi$ and their partial derivatives, but it seems a bit messy to write these in terms of $\phi$ etc. What I've tried:

Writing $\phi = u + iv$ we get

$$ \begin{eqnarray} \iint_{S_r} dx\, dy &=& \int_{\partial\phi(B_1\setminus B_r)}x\ dy \\ &=& \int_{\partial B_1- \partial B_r} uv_x \ dx + uv_y \ dy \end{eqnarray} $$

If I go on from here by substituting $u = \frac12(\phi + \overline \phi), v = \frac12(\phi - \overline \phi)$, then the result I come up with for the integral over $\partial B_r(0)$ seems to be

$$\pi\left(\frac{-1}{r^2} + \sum_{n=1}^\infty n|a_n|^2 r^{2n}\right)$$

But I'm neither really sure whether this is right (I'm guessing not, because the series has the wrong form), nor do I know what to do with that.

So my question really is:

  • What is the best approach to this problem? Any hints/solutions would be appreciated.
  • What is the motivation behind considering $S_r$ (probably a correct result would answer this retrospectively)?

Thanks for your thoughts and comments.

  • $\begingroup$ I suspect an error in the problem, because there is a very similar proof in Carleson and Gamelin's "Complex dynamics", p.2. The theorem they state is with the sum of the squares of modulus of $a_n$ just like what you obtain $\endgroup$
    – Albert
    Dec 6, 2011 at 19:38
  • $\begingroup$ @Glougloubarbaki: Thanks for the reference. I've had a look at it and it seems that you are right. $\endgroup$
    – Sam
    Dec 6, 2011 at 20:02
  • $\begingroup$ @Glougloubarbaki: I have now also found it in Rudin's Real and Complex Analysis (actually the entire problem) and you are definitely right. If you want to write your comment as an answer, then I will accept it. $\endgroup$
    – Sam
    Dec 6, 2011 at 20:52

1 Answer 1


This question is answered in Rudin's Real and Complex Analysis, Ch. 14 "Conformal Mapping", p. 285 onwards in the third edition of the book.


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