2
$\begingroup$

This is for homework, so just hints please! The question asks

If $f$ is a meromorphic function in $\mathbb{C}$ that satisfies $|f(z) z^2| \leq 1$ for $|z| \geq 1$, then evaluate $\int_{\partial \mathbf{D}} f(z) dz$ (where $\mathbf{D}$ represents the unit disk).

Since $f$ is meromorphic on $\mathbb{C}$, it is certainly meromorphic on $\mathbf{D}$. This means that there are at most finitely many poles $\{ z_0, \dotsc, z_n \}$ of $f$ in $\mathbf{D}$. By the residue theorem, then, $$ \int_{\partial \mathbf{D}} f(z) dz = \sum_{i=0}^n \text{Res}_{z = z_i} \left[ f(z) \right]. $$ However, by the condition $|f(z) z^2| \leq 1$ for $|z| \geq 1$, we see that the Laurent series of $f$ about each $z_i$ contains no powers greater than $-2$. For example, $f(z) = \frac{1}{z} + \frac{1}{z^3}$ does not fit our criterion. The residue of $f$ at a point is defined to be the coefficient of $\frac{1}{z}$ in the Laurent expansion of $f$. Thus, $\text{Res}_{z = z_i} \left[ f(z) \right] = 0$ for every $i$, and hence $$ \int_{\partial \mathbf{D}} f(z) dz = 0. $$

Does this look OK? I have a feeling that there is a mistake somewhere but am not sure where.

$\endgroup$
6
  • $\begingroup$ Looks good to me. $\endgroup$ – copper.hat Nov 20 '13 at 1:46
  • $\begingroup$ I agree. Looks good. $\endgroup$ – mathematician Nov 20 '13 at 1:51
  • $\begingroup$ Looks good. Alternative proof to check your argument: Deform the integration path from $\partial D$ to the boundary of a really large circle and estimate the integral. $\endgroup$ – Hans Engler Nov 20 '13 at 1:51
  • $\begingroup$ Can you perhaps be more explicit about how you are reaching the conclusion that the Laurent coefficients must be zero for $n > -2$? $\endgroup$ – Eric Auld Nov 20 '13 at 1:59
  • $\begingroup$ @EricAuld If a coefficient of the Laurent series of $f$ is nonzero for $n > 2$, then $|f(z) z^2| \nleq 1$ for $|z| \geq 1$, right? $\endgroup$ – tylerc0816 Nov 20 '13 at 2:02
2
$\begingroup$

The end result is correct. However, the way to it isn't.

Thus, $\text{Res}_{z = z_i} \left[ f(z) \right] = 0$ for every $i$

is wrong. Consider for example

$$f(z) = \frac12\left(\frac{1}{z-\frac12} - \frac{1}{z+\frac12}\right) = \frac{1}{2(z^2-\frac14)}.$$

That satisfies the criterion $\lvert f(z)z^2\rvert \leqslant 1$ for $\lvert z\rvert \geqslant 1$, but it has two poles with nonzero residue.

The condition $\lvert f(z)z^2 \rvert \leqslant 1$ for $\lvert z\rvert \geqslant 1$ ensures that $f$ has a zero of order (at least) $2$ in $\infty$. Thus $f$ is a rational function, and the sum of all residues of a rational function is $0$. The fact that $f$ has a multiple zero in $\infty$ means that the residue in $\infty$ is $0$, hence the sum of all residues in $\mathbb{C}$ (which is the sum of the residues in $\mathbb{D}$) is $0$.

$\endgroup$
8
  • $\begingroup$ More specifically, the sum of the residues of a rational function $\frac{p(z)}{q(z)}$ -- such that $\deg p(z) < \deg q(z) - 1$ (which is what we have) -- is 0, correct? $\endgroup$ – tylerc0816 Nov 20 '13 at 2:07
  • $\begingroup$ Any rational function, but you must include the residue in $\infty$. $\endgroup$ – Daniel Fischer Nov 20 '13 at 2:09
  • $\begingroup$ Why is the sum of all residues of a rational function equal to zero? If it is lengthy to explain, do you know a reference? Thanks $\endgroup$ – Eric Auld Nov 25 '13 at 2:33
  • $\begingroup$ @EricAuld The short reason is that the sum of residues of a meromorphic function on a compact Riemann surface is $0$, and a rational function is a meromorphic function on the compact Riemann surface $\widehat{\mathbb{C}}$. One can of course prove that special case with more elementary means, but I'll have to think about it, need to fully wake up first. $\endgroup$ – Daniel Fischer Nov 25 '13 at 10:28
  • $\begingroup$ @DanielFischer Did you ever get a chance to think about this one? Thanks $\endgroup$ – Eric Auld Dec 6 '13 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.