In Ahlfors' complex analysis text, page 321 he discusses the function $$p(z)=\frac{f_1 f_2''-f_2 f_1''}{f_1 f_2'-f_2 f_1'} \tag{24} $$ where $f_1,f_2$ are analytic in some simply connected region $\Omega$, which doesn't contain the points $0,1$, and they can be continued analytically along any arc which avoids the points $0$ and $1$. He gets the decomposition $$p(z)=\frac{\alpha_1+\alpha_2-1}{z}+\frac{\beta_1+\beta_2-1}{z-1}+p_0(z) $$ where $p_0$ is free from poles at $0$ and $1$. He further says

According to its definition $(24)$, $p(z)$ is the logarithmic derivative of an entire function; as such it has, in the finite plane, only simple poles with positive integers as residues. Moreover, the development of $p(z)$ at $\infty$ must begin with the term $-(\gamma_1 + \gamma_2 + 1)/z$. Hence $p(z)$ has only finitely many poles, and their residues must add up to $-(\gamma_1 + \gamma_2 + 1)$.

I have several question regarding this:

  1. Why is the function $f_1 f_2'-f_2 f_1'$ entire as he claims? What about the points $0,1$?

  2. Why $p(z)$ is bound to have only finitely many poles?

  3. Why is the sum of residues in the plane must equal the residue at $\infty$?

Thank you!


2 Answers 2


The last chapter of Ahlfors contains numerous errors. One could only guess that not so many people actually proof-read it.

Any way, the claim that $𝑝(𝑧)$ is the logarithmic derivative of an entire function is obviously wrong. $𝑝(𝑧)$ is however the logarithmic derivative of an analytic function defined on $\mathbb{C} - \{0, 1\}$.

This means that $p(z)$ has only simple poles with positive integers as residues on $\mathbb{C} - \{0, 1\}$. The word positive is important here.

Furthermore, the decomposition shows that $p(z)$ has simple poles at 0 and 1. So the residual theorem states that their residues (including those of the poles at 0 and 1) must add up to $-(\gamma_1 + \gamma_2 + 1)$. But the two poles at 0 and 1 already give us $(\alpha_1 + \alpha_2 - 1) + (\beta_1 + \beta_2 - 1) = -(\gamma_1 + \gamma_2 + 1)$. So there cannot be any other poles, whose residuals would have to be positive integers.


I haven't read Ahlfors' book but I think I can maybe answer your questions anyway:

  1. It's not an entire function, it has poles at $0$ and $1$ (check your claim that Ahlfors says it is entire)
  2. I think this is because there is a good old pole at $\infty$. Note that poles are always isolated (an analytic function always has no other singularities in a neighborhood of a pole). In particular the set of poles is always discrete. Moreover, if your function has no essential singularities (which I'm assuming is the case here, although I don't have enough info to back up that claim), then the set of poles is also closed (because an accumulation point could only be an essential singularity). Since the Riemann sphere $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ is compact, your set of poles must be finite.
  3. This is a consequence of the residue theorem. Let $C = \partial D(0,R)$ be a circle with $R>0$ large enough so that $f$has no singularities on $\{z\in \mathbb{C}, |z|>R\}$. Define the residue at $\infty$ as ${1 \over 2i\pi} \int_C f(z)\,dz$ where $C$ is positively oriented around $\infty$, $\it{i.e.}$ $C$ is negatively oriented (around the origin). Apply the residue theorem (is that convincing enough?)

Hope I didn't say anything wrong!

Edit: Actually, the way I wrote 2. is slightly dubious, let me rectify. Usually in complex analysis, you always consider functions with isolated singularities only (so I'm assuming this is the case for your functions here). So the set of singularities is always discrete and closed (because an accumulation point would be a non-isolated singularity).

  • $\begingroup$ Sorry for the late response. I have several questions: 1. Why is it not entire? How can I see there are poles? 3. I think the identity function $f(z)=z$ is a counterexample. It has residue 0 in the finite plane, and a simple pole (with residue 1?) at $\infty$? Thanks $\endgroup$
    – user1337
    Feb 9, 2014 at 18:09
  • $\begingroup$ 1. Sorry, I meant to say that $p$ has poles at $0$ and $1$ (just lookking at its expression). Anyway, you do not give enough information about $f_1$, $f_2$ to derive much of anything. 3. No it's not. Just use the definition I gave, you'll see that the residue at infinity is $0$. I agree that this is confusing because $\infty$ is a simple pole of $z \mapsto z$ indeed, but the explanation is that we shouldn't really talk about residues of a function $f$ but rather of the $1$-form $f(z)\,dz$. $\endgroup$
    – Seub
    Feb 9, 2014 at 18:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .