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Let , $f$ be analytic for $|z|<2$. Show that $$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz=\begin{cases}\overline{f(0)} &\text{ if } |a|<1\\\overline{f(0)}-\overline{f(1/a)}&\text{ if }|a|>1\end{cases}$$

By putting $z=e^{i\theta}$ , $\displaystyle\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz=\int_0^{2\pi}\frac{\overline{f(e^{i\theta})}}{e^{i\theta}-a}ie^{i\theta}\,d\theta=i\int_0^{2\pi}\frac{\overline{f\left(\overline{e^{-i\theta}}\right)}}{1-ae^{-i\theta}}\,d\theta$.

Now putting , $e^{-i\theta}=t$ , it becomes $\displaystyle =-\int_{|t|=1}\frac{\overline{f(\bar t)}}{1-at}\,dt$. As $f(t)$ is analytic so $\overline{f(\bar t)}$ is also analytic. Now we can apply the residue theorem and finally I got the desire integral $=-\overline{f(0)}$ when $|a|<1$.

My problem is about the sign. I got an extra negative sign. Please verify my proof & detect where I made mistake ...

Empty
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  • When I do the usual substitution $z = e^{it}$, then substitute $s = -t$ and then go back, I get that the integral is equal to $\frac 1 a\int_{|z|=1}\frac{h(z)}{z},dz$, where$$h(z) = \frac{\overline{f(\overline z)}}{a^{-1}-z}.$$ Hence, if $|a| < 1$, $h$ is analytic in a neighborhood of the closed unit disc and so, the original expression is equal to $h(0)/a = \overline{f(0)}$. – Friedrich Philipp Mar 04 '16 at 19:06

1 Answers1

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$$\begin{align} \oint_{|z|=1}\frac{\overline{ f(z)}}{z-a}\,dz&=\int_0^{2\pi}\frac{\overline{ f(e^{i\theta})}}{e^{i\theta}-a}ie^{i\theta}\,d\theta \tag 1\\\\ &=\int_{0}^{-2\pi}\frac{\overline{ f(e^{-i\theta})}}{e^{-i\theta}-a}ie^{-i\theta}\,(-1)\,d\theta \tag 2\\\\ &=\int_{-2\pi}^{0}\frac{\overline{ f(\overline{e^{i\theta}})}}{\left(e^{-i\theta}-a\right)e^{i2\theta}}ie^{i\theta}\,d\theta \tag 3\\\\ &=-\int_0^{2\pi}\frac{\overline{ f(\overline{e^{i\theta}})}}{\left(ae^{i\theta}-1\right)e^{i\theta}}ie^{i\theta}\,d\theta \tag 4\\\\ &=-\oint_{|z|=1}\frac{\overline{f(\overline{z})}}{z(az-1)}\,dz \tag 5\\\\ &=\oint_{|z|=1}\overline{f(\overline{z})}\left(\frac{1}{z}-\frac{a}{az-1}\right)\,dz \tag 6\\\\ &=2\pi i \begin{cases} \overline{f(0)}&,|a|<1\\\\ \overline{f(0)}-\overline{f(\overline{1/a})}&,|a|>1 \tag 7 \end{cases} \end{align}$$

In arriving at $(1)$, we parametrized the unit circle letting $z=e^{i\theta}$.

In going from $(1)$ to $(2)$, we enforced the substitution $\theta \to -\theta$.

In going from $(2)$ to $(3)$, we absorbed the factor of $-1$ by transposing the integration limits. We wrote $e^{-i\theta}=\overline{e^{i\theta}}$ in the argument of $f$. And we multiplied the numerator and denominator by $e^{i2\theta}$.

In going from $(3)$ to $(4)$, we multiplied $\left(e^{-i\theta}-a\right)e^{i\theta}=-(ae^{i\theta}-1)$ and then exploited the $2\pi$-periodicity of the integrand to transform the limits from $-2\pi$ to $0$ to $0$ to $2\pi$.

In going from $(4)$ to $(5)$, we moved from a parametric description back to the contour integral representation.

In going from $(5)$ to $(6)$, we used partial fraction expansion.

And in going from $(6)$ to $(7)$, we used the residue theorem.

Mark Viola
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  • Yes..I realize my mistake..But I can't understand your 3rd line to 4th line. How $\displaystyle \int_{-2\pi}^0=-\int_0^{2\pi}$ ? – Empty Mar 05 '16 at 03:23
  • The integrand is $2\pi$-periodic – Mark Viola Mar 05 '16 at 07:14
  • From 1st line to 2nd line you put $z=e^{i\theta}$. But how you get from 2nd line to 3rd ? How $\int_0^{2\pi}$ transformed into $\int_0^{-2\pi}$ ? – Empty Apr 17 '16 at 18:50
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    From the second to third line, we absorbed the factor $(-1)$ by transposing the limits of integration. From the first to the second line, we enforced the substitution $\theta \to -\theta$. – Mark Viola Apr 17 '16 at 19:22
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    I've edited to provide a line-by-line explanation of the steps in the development. Again, I really want to give you the best answer I can. So, please just let me know how I can improve my answer. -Mark – Mark Viola Apr 18 '16 at 04:12
  • You're welcome! My pleasure. -Mark – Mark Viola Apr 18 '16 at 14:23
  • @MarkViola Is there a way for one get from step 6 to 7 without use of the residue theorem? –  Nov 18 '18 at 18:52
  • @mathwolf What did you have in mind. – Mark Viola Nov 18 '18 at 19:04
  • @MarkViola I've been given this Q and in a previous part of the Q I proved that for a continuous function $\phi:A={z\in\mathbb{C}:|z|=1}\rightarrow\mathbb{C}$ and $\gamma$ as above, $$\overline{\int_{\gamma}\phi(z),, dz} = -\int_{\gamma} \overline{\phi(z)},,\frac{dz}{z^2}$$ Then, using this (and CIF) I've managed to get the following: $$\overline{\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z}dz},,-,,\overline{\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-1/\bar{a}}dz}$$ which by CIF is equal to $\overline{f(0)}-\overline{f(1/\bar{a})}$ . Now I just don't know how to get the other case. –  Nov 18 '18 at 19:23
  • @MarkViola I ask for a method without the residue theorem as we have yet to cover it. –  Nov 18 '18 at 19:24
  • From CIF, the second integral is $0$ for $|a|<1$. – Mark Viola Nov 18 '18 at 19:36
  • @MarkViola So, CIF states that: Let U be a domain. Let $\gamma$ be a positively oriented, simple closed contour with its image and interior lying entirely within U. Suppose that $a$ is a point in the interior of $\gamma$. If f is holomorphic on U, then $$f(a) = \frac{1}{2\pi i} \int_{\gamma}\frac{f(z)}{z-a},dz$$ I understand that as $|a|<1,, ,1/|\bar{a}| >1$ then $1/\bar{a}$ is not in the interior of $\gamma$ which means CIF can't be applied. But I don't understand how this translates to the integral being $0$. What am I missing here? –  Nov 18 '18 at 19:54
  • @MarkViola Just following up in case you didn't get a notification for my previous comment. –  Nov 19 '18 at 08:49
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    @mathwolf If $|a|>1$he integrand is analytic for $|z|\le 1$. Cauchy's Integral Theorem guarantees that the integral over $\gamma$ is zero. – Mark Viola Nov 19 '18 at 13:49
  • @MarkViola Thank you! –  Nov 19 '18 at 15:00
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    You're welcome. My pleasure. – Mark Viola Nov 19 '18 at 18:22