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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.

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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 $\zeta$.

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Aha! Thanks. That makes so much sense. – Daniel Briggs May 25 '11 at 8:24

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