Let $\Gamma$ denote the positively oriented circle of radius $2$ with center at the origin. Let $f$ be an analytic function on $\{z\in\mathbb{C}\ |\ |z| \gt 1\},$ and let $$\lim_{z\to\infty} f(z)=0.$$ Prove that $$f(z)=\dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{f(\zeta)}{z-\zeta}\ d\zeta$$ for all $z\in\mathbb{C}$ with $|z|>2.$

By the given condition it is clear that $\lim\limits_{z \to 0} f \left ( \dfrac 1 z \right ) = 0.$

Now define a function $g : B(0,1) \longrightarrow \Bbb C$ by

$$ g(z) = \left\{ \begin{array}{ll} f \left (\dfrac 1 z \right ) & \quad 0 \lt |z| \lt 1 \\ 0 & \quad z = 0 \end{array} \right. $$

Then $g$ is analytic on $B(0,1).$ Let $\Gamma' (t) = \frac {1} {\Gamma (t)},$ where $t$ varies over the parameter interval of $\Gamma.$ Then $\Gamma'$ is a circle of radius $\frac {1} {2}$ traversing in the clockwise direction. Since $g$ is analytic on and inside of $\Gamma'.$ So by Cauchy's integral theorem it follows that for all $z \in \Bbb C$ with $|z| \lt \frac {1} {2}$ we have $$g(z) = - \dfrac {1} {2 \pi i} \int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - z}\ d\zeta.$$

Hence for all $z \in \Bbb C$ with $|z| \gt 2$ we have $$f(z) = g \left ( \frac 1 z \right ) = - \dfrac {1} {2 \pi i} \int_{\Gamma'} \dfrac { g (\zeta) } {\zeta - \frac 1 z}\ d\zeta.$$

Can it be shown that $$-\displaystyle { \int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - \frac 1 z}\ d\zeta = \int_{\Gamma} \dfrac {f(\zeta)} {z - \zeta}\ d\zeta}$$ for all $z \in \Bbb C$ with $|z| \gt 2\ $?

Any help in this regard will be highly appreciated. Thanks in advance.

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    $\begingroup$ Please have a look at the question @Surb. It is clearly mentioned in the question that $\Gamma$ is a positively oriented circle of radius $2$ with center at the origin. $\endgroup$ – Anacardium Sep 10 '20 at 7:35
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    $\begingroup$ After a bit of calculation I end up with the following equality. $$-\int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - \frac {1} {z}}\ d\zeta = \int_{\Gamma} \dfrac {f(\zeta)} {\zeta}\ d\zeta + \int_{\Gamma} \dfrac {f(\zeta} {z - \zeta}\ d\zeta.$$ $\endgroup$ – Anacardium Sep 10 '20 at 7:40

$$-\int_{\Gamma'}\frac{g(t)}{t-\frac1z}dt=^1\int_\Gamma \frac{f(\zeta)}{\frac1\zeta-\frac1z}\frac{d\zeta}{\zeta^2}=\int_\Gamma\frac{f(\zeta)}{z-\zeta}\frac{z}{\zeta}d\zeta=\\=\int_\Gamma \frac{f(\zeta)(z-\zeta)}{(z-\zeta)\zeta}+\frac{f(\zeta)}{z-\zeta}d\zeta=\int_\Gamma \frac{f(\zeta)}{z-\zeta}d\zeta+\int_\Gamma \frac{f(\zeta)}{\zeta}d\zeta$$ We now have to prove that the second term in the RHS is $0$: $$ \int_\Gamma \frac{f(\zeta)}{\zeta}d\zeta=^1-\int_{\Gamma'}\frac{g(t)}{t} dt=g(0)=0$$

The result follows.

If you know CIT for a chain homologous to zero, one can prove the result in a straightforward manner: let $\gamma$ be the circle of radius $2$ (positively or.) and $\gamma_R$ the (neg. or.) circle of radius $R$. Then, for $2<|z|<R$

$$f(z)=\frac1{2\pi i}\left(\int_{\gamma}\frac{f(\zeta)}{z-\zeta}d\zeta-\int_{\gamma_R}\frac{f(\zeta)}{z-\zeta}d\zeta\right)\\ \lim_{R\to \infty}\left|\int_{\gamma_R}\frac{f(\zeta)}{z-\zeta}d\zeta\right| \le\lim_{R\to \infty}\text{max}_{\gamma_R}|f|\frac{2\pi R}{R-|z|}=0$$

The result follows. $^1$: apply the change of variable $t=\frac{1}{\zeta}, dt=-\frac{d\zeta}{\zeta^2}$

  • $\begingroup$ The variable $t$ is quite awkward looking. As it is often referred to as real variable. Changing the complex variable $\zeta$ to let's say $\xi$ may be better to use, I guess. $\endgroup$ – Anacardium Sep 10 '20 at 7:56
  • $\begingroup$ @Anacardium de gustibus non est disputandum $\endgroup$ – Caffeine Sep 10 '20 at 8:01
  • $\begingroup$ What? Is it Latin? $\endgroup$ – Anacardium Sep 10 '20 at 8:06
  • $\begingroup$ It turns out from what you prove as an alternative way that for any $z \in \Bbb C$ the equality holds since you tend $R$ to $\infty.$ Right? $\endgroup$ – Anacardium Sep 10 '20 at 8:16
  • $\begingroup$ In order for the equality to hold, $|z|>2$. Apart from that, the equality holds without an upper bound exactly because $R\to \infty$, as you noticed $\endgroup$ – Caffeine Sep 10 '20 at 8:18

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