Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In studying early for qualification exams, I came across the following problem (UMass Amherst Graduate Qualifying Exams / Fall 2010 Complex Analysis Exam (see #10)):

Let $D$ denote the open set $D:=\{z:|z|>1\}$ and $\bar D:=\{z:|z|\geq1\}$ its closure. Suppose $f(z)$ is holomorphic on an open set $U$ containing $\bar D$ and that $$\lim_{z\to\infty}f(z)=1.$$ Show that for any $z$ in $D,$ $$\int_{|\zeta|=1}\frac{f(\zeta)}{\zeta-z}d\zeta=2\pi i(1-f(z)).$$

Note that D here means the opposite of what it usually means. What I am interested in is where my approach goes wrong.

Let $g(\zeta)=f(1/\zeta).$ Then


The reasoning is that the values on the right-hand side are identical to those on the left, and the direction is just reversed. So we would like to prove that

$$\int_{|\zeta|=1}\frac{g(\zeta)}{\zeta^{-1}-z}d\zeta=2\pi i(f(z)-1),$$

but for some reason I keep getting that it's $-2\pi if(z)/z^2.$ The nice thing about $g$ is that it's holomorphic within an open set containing the unit disk, since the limit condition on $f$ allows us to define $g(0)=1.$ Thus, the integrand has only one pole, of order 1, at $w=1/z.$ Then $$\mathrm{Res}_w=\lim_{\zeta\to w}\frac{g(\zeta)}{\zeta^{-1}-z}\left(\zeta-\frac{1}{z}\right)$$ $$=\lim_{\zeta\to w}\frac{\zeta g(\zeta)}{1-z\zeta}\frac{\zeta z-1}{z}$$ $$=-\frac{f(z)}{z^2}.$$

Thus either I'm wrong, the question's wrong, or the function $f(z)$ has to be $z^2/(1-z^2).$ I would imagine it's the former.

Thanks for your help.

share|cite|improve this question
up vote 6 down vote accepted

Your equation


is wrong -- substituting $u=\zeta^{-1}$ yields

$$\int_{|\zeta|=1}\frac{f(\zeta)}{\zeta-z}d\zeta=-\int_{|u|=1}\frac{g(u)}{u^{-1}-z}\left(-\frac{1}{u^2}\right)d u\;.$$

Your argument that "the direction is just reversed" doesn't work, since the change in $\zeta^{-1}$ isn't simply minus the change in $\zeta$.

share|cite|improve this answer
Aha! Thanks. That makes so much sense. – Daniel Briggs May 25 '11 at 8:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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