# Proving that $\frac{1}{2 \pi i}\oint_{\gamma_{1}} \frac{d \zeta}{(\zeta - 1)\zeta + 1}$ is equal to itself?

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:

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

• Sorry but $\gamma_1=\gamma_2$? Another strange thing is that you seem to consider integrals on disks, which are simply undefined. – Did May 28 '17 at 16:21
• @upvoters Care to explain what the question is about? – Did May 28 '17 at 16:23
• Any operations involving Cauchy's Integral Formula involve it being a closed disk since that's how it's defined. – Zophikel May 28 '17 at 16:38
• Not at all. Please check your notes, which should mention integrals along circles, say, but never on disks. – Did May 28 '17 at 16:39
• My mistake it would involve the unit disk – Zophikel May 28 '17 at 17:19

$$\frac{1}{2 \pi i}\oint_{\Gamma_{1} = {|\zeta - (-1)| = 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta - (-1)| =1}\frac{\frac{1}{\zeta - 1}}{\zeta -(-1) }d \zeta = \frac{1}{-1-1}=-\frac{1}{2}$$
$$\frac{1}{2 \pi i}\oint_{\Gamma_{2} = {|\zeta - 1| = 1}} \frac{d \zeta}{(\zeta - 1)(\zeta + 1)}= \frac{1}{2 \pi i} \oint_{\Gamma_{1}=|\zeta -1| = 1}\frac{\frac{1}{\zeta + 1}}{\zeta -1 }d \zeta = \frac{1}{1+1}=\frac{1}{2}$$
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.
• The ending should actually just be $f(1)$, not $2\pi if(1)$. – Simply Beautiful Art May 28 '17 at 18:06