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In the text "Functions of a Complex Variable" I'm having trouble verifying if my proof of $(0.)$

$(0.)$

$$\frac{1}{2 \pi i}\oint_{\gamma_{1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)} = \frac{1}{2 \pi i}\oint_{\gamma_{2}}\frac{d \zeta}{(\zeta - 1)(\zeta + 1)}$$

Remark: $\gamma_{1}$ is $\partial{D(1,1)}$ with a clockwise orientation and where $\gamma_{2}$ is $\partial{D}(-1,1)$ equipped with a counterclockwise orientation.

Proposition(1.1):(Cauchy's Integral Formula)

Suppose $U$ is an open subset of $\mathbb{C}$, $f : U \rightarrow \mathbb{C}$ is a holomorphic the unit disk ${D} = {\zeta : | \zeta - P | \leq r}$ is completely contained in $U$. Let $\Gamma$ be the circle forming the boundary of ${D}$. Then for every a in the interior of ${D}$

$$f(a)=\frac{1}{2 \pi i}\int_{\Gamma}\frac{f(z)}{z-a}dz$$

Applying Cauchy's Integral Formula to the RHS and LHS side of $(0.)$, one can make the following assertions:

$$\frac{1}{2 \pi i}\oint_{\Gamma_{1} = {|\zeta - (-1)| \leq 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta - (-1)| \leq 1}\frac{\frac{1}{\zeta + 1}}{\zeta -1 }d \zeta = 2 \pi i \cdot f(1)$$

$$\frac{1}{2 \pi i}\oint_{\Gamma_{2} = {|\zeta - (-1)| \leq 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta - (-1)| \leq 1}\frac{\frac{1}{\zeta + 1}}{\zeta -1 }d \zeta = 2 \pi i \cdot f(1)$$

Obviously accounting on the far RHS side of our previous observations it easy to notion that:

$$2 \pi i \cdot f(1) = 2 \pi i \cdot f(1)$$

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  • $\begingroup$ Sorry but $\gamma_1=\gamma_2$? Another strange thing is that you seem to consider integrals on disks, which are simply undefined. $\endgroup$ – Did May 28 '17 at 16:21
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    $\begingroup$ @upvoters Care to explain what the question is about? $\endgroup$ – Did May 28 '17 at 16:23
  • $\begingroup$ Any operations involving Cauchy's Integral Formula involve it being a closed disk since that's how it's defined. $\endgroup$ – Zophikel May 28 '17 at 16:38
  • $\begingroup$ Not at all. Please check your notes, which should mention integrals along circles, say, but never on disks. $\endgroup$ – Did May 28 '17 at 16:39
  • $\begingroup$ My mistake it would involve the unit disk $\endgroup$ – Zophikel May 28 '17 at 17:19
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First: as Did says, it doesn't make sense to integrate on disks in the complex plane, you integrate on circles. That being said, there are some errors in your last lines, they should read

$$\frac{1}{2 \pi i}\oint_{\Gamma_{1} = {|\zeta - (-1)| = 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta - (-1)| =1}\frac{\frac{1}{\zeta - 1}}{\zeta -(-1) }d \zeta = \frac{1}{-1-1}=-\frac{1}{2}$$

$$\frac{1}{2 \pi i}\oint_{\Gamma_{2} = {|\zeta - 1| = 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta -1| = 1}\frac{\frac{1}{\zeta + 1}}{\zeta -1 }d \zeta = \frac{1}{1+1}=\frac{1}{2}$$

And you should change the sign of the second integral because of the orientation. In the two cases the $a$ in Cauchy's formula is the center of the circle.

Note that the integration is done on the two circles which are the boundaries of the disks you were talking about (I changed the domain of integration in the integrals). If this is not clear to you I suggest you re read the chapters about integration in the complex plane in your textbook.

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    $\begingroup$ The ending should actually just be $f(1)$, not $2\pi if(1)$. $\endgroup$ – Simply Beautiful Art May 28 '17 at 18:06
  • $\begingroup$ I should get more sleep $\endgroup$ – user438666 May 28 '17 at 18:11

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