### Author Topic: Q4 TUT 0301  (Read 2782 times)

#### Victor Ivrii ##### Q4 TUT 0301
« on: October 26, 2018, 05:54:27 PM »
Evaluate the given integral using Cauchy’s Formula or Theorem. Orientation counter-clockwise:
$$\int_{|z+1|=2} \frac{z^2\,dz} {4-z^2}.$$

#### Meng Wu ##### Re: Q4 TUT 0301
« Reply #1 on: October 26, 2018, 05:54:48 PM »
$|z+1|=2 \text{ is the circle at point -1 with radius }2.$
$$\int_{|z+1|=2}\frac{z^2}{4-z^2}dz=\int_{|z+1|=2}\frac{z^2}{(2-z)(2+z)}dz$$
Let $p=2,q=-2 \text{ where } p \text{ lies outside of the circle |z+1|=2}.$
\begin{align}f(z)=\frac{1}{2\pi i}\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi \\ \Rightarrow \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi&=2\pi i f(z)\end{align}
$\text{Where } f(\xi)=\frac{\xi^2}{2-\xi}, \int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=\int_{|z+1|=2}\frac{\frac{\xi^2}{2-\xi}}{\xi-(-2)}d\xi.$ $\\$ $f(z=-2)=\frac{(-2)^2}{2-(-2)}=1.$ $\\$
Therefore, $$\int_{|z+1|=2}\frac{f(\xi)}{\xi-z}d\xi=2\pi i f(z)=2\pi i\cdot 1=2\pi i$$
« Last Edit: October 26, 2018, 05:58:46 PM by Meng Wu »

#### Yuechen Huang

• Jr. Member
•  • Posts: 6
• Karma: 2 ##### Re: Q4 TUT 0301
« Reply #2 on: October 26, 2018, 06:18:30 PM »
First we have
\begin{equation}
\int\limits_{\mid z+1 \mid = 2}\frac{z^2}{4-z^2}\,dz = \int\limits_{\mid z+1 \mid = 2}\frac{z^2}{(2+z)(2-z)}\,dz
\end{equation}
Point $z=2$ is outside of the circle $\mid z+1 \mid = 2$ and Point $z=-2$ is inside of the circle $\mid z+1 \mid = 2$
\begin{equation}
\int\limits_{\mid z+1 \mid = 2}\frac{z^2/(z-2)}{(z+2)}\,dz = \int\limits_{\mid z+1 \mid = 2}\frac{z^2/(z-2)}{z-(-2)}\,dz
\end{equation}
This gives us function $f(z)$
\begin{equation}
f(z) = \frac{z^2}{2-z} \Rightarrow f(-2) = \frac{(-2)^2}{2-(-2)} = \frac{4}{4} = 1
\end{equation}
So Cauchy's Formula gives us
\begin{equation}
\int\limits_{\mid z+1 \mid = 2}\frac{z^2}{4-z^2}\,dz = 2\pi if(-2) = 2\pi i
\end{equation}