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.

  • $\begingroup$ 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

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 $0$.

  • $\begingroup$ 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

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