Prove that $f(z)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{z-\zeta}\ d\zeta,$ for all $z\in\mathbb{C}$ with $|z| \gt 2.$ [duplicate]

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.

• 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. – Anacardium Sep 10 '20 at 7:35
• After a bit of calculation I end up with the following equality. $$-\int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - \frac {1} {z}}\ d\zeta = \int_{\Gamma} \dfrac {f(\zeta)} {\zeta}\ d\zeta + \int_{\Gamma} \dfrac {f(\zeta} {z - \zeta}\ d\zeta.$$ – 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|

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

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

• 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. – Anacardium Sep 10 '20 at 7:56
• @Anacardium de gustibus non est disputandum – Caffeine Sep 10 '20 at 8:01
• What? Is it Latin? – Anacardium Sep 10 '20 at 8:06
• 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? – Anacardium Sep 10 '20 at 8:16
• 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 – Caffeine Sep 10 '20 at 8:18