Context: This is the Laurent Series section of my text, which can be found here on pages 129-131. It is right after he derived the expression for Laurent coefficients for a function that is analytic in an annulus region, $R_1 < |z-z_0| < R_2$, to be $C_n = \frac{1}{2\pi i}\oint \frac{f(z)}{(z-z_0)^{n+1}}dz$, getting $f(z) = \sum_{-\infty}^{\infty}C_n(z-z_0)^n$.

After showing this, he discusses two different cases:

1.) Suppose $f(z)$ is analytic everywhere inside $R_1$, then $C_n = 0$ $\forall$ $n \leqslant -1$. This is clear to me, since for this case the integrand is analytic and hence must be zero by Cauchy's Theorem.

2.) This is the one I do not see. Suppose $f(z)$ is analytic everywhere outside $R_2$, then $C_n = 0$ $\forall$ $n > 0$. He does not comment on this result, it is just stated, which makes me think that it is something obvious that I am missing. Any contour $C$ that encloses $|z-z_0| = R_1$ still obviously encloses singular points, so our handy integration tricks cannot be applied (or so it seems). How exactly is this done? Apologies if this is abundantly clear.

Edit: On my first read through of this part of the section, I was fundamentally confused about what the author was doing. It is now clear that the answer to this question is given directly after it is mentioned, which goes like this: deform the contour to a larger one of radius $R$, employ the transformation $z = 1/t$, use boundedness of $f(1/t)$ for small $t$ from $f$ being analytic at infinity from the assumption, and use the reverse triangle inequality to show that $|C_{n}| \leq 2M\epsilon^n \rightarrow 0$ as $\epsilon \rightarrow 0$, where $\epsilon = 1/R$ is the contour radius.

  • $\begingroup$ Hint: consider $\displaystyle f\bigg( \frac1{z-z_0}+z_0 \bigg)$. $\endgroup$ – Greg Martin Mar 17 '14 at 6:12
  • $\begingroup$ @GregMartin I am not quite sure what you are getting at. I see that it eliminates the $z_0$ in the series representation, but how does that help with the integration? Additionally, how do we know that the function is even defined at this point? For example, let $z_0 = 0$ which reduces this to $f(\frac{1}{z})$, so if $|1/z| < R_1$ then we know nothing of the existence of this functional value. $\endgroup$ – dsm Mar 17 '14 at 15:04
  • $\begingroup$ I agree it's easier to look at the case $z_0=0$. And you're right, I think: the function $f(z)=z+1/z$ is analytic outside $\{|z|=R_2\}$ for any $R_2$, yet has a positive Laurent coefficient $C_1$. Unless, when the author says "analytic everywhere outside $R_2$", they actually mean analytic at infinity as well. Is that possibly the intended interpretation? $\endgroup$ – Greg Martin Mar 17 '14 at 20:51
  • $\begingroup$ @GregMartin From his statements I certainly assume that the point at infinity is included. I just supposed $z_0 = 0$ to shed light on my confusion with your original evaluation, as I still want to figure out the general case. $\endgroup$ – dsm Mar 18 '14 at 3:37
  • 1
    $\begingroup$ @Daniel I am looking at the text, and I see that the author proves the result you are asking about. (Starting from "In particular..." to middle of pg. 131.) So I am a bit confused about your question: Are you asking for clarification of his computation? $\endgroup$ – Braindead Mar 24 '14 at 3:11

Your Answer

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

Browse other questions tagged or ask your own question.