### Author Topic: TT2A Problem 1  (Read 1613 times)

#### Victor Ivrii ##### TT2A Problem 1
« on: November 24, 2018, 04:54:06 AM »
Using Cauchy's integral formula calculate
$$\int_\Gamma \frac{z\,dz}{z^2-4z+5},$$
where $\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter
of $2\pm i$ in the following cases

(a) The point $2+i$ is inside  $\Gamma$ and $2-i$ is outside  it;

(b) The point $2-i$ is inside  $\Gamma$ and $2+i$ is outside it;

(c) Both points $2\pm i$ are inside  $\Gamma$.

#### ZhenDi Pan

• Jr. Member
•  • Posts: 10
• Karma: 20 ##### Re: TT2A Problem 1
« Reply #1 on: November 24, 2018, 05:30:25 AM »
We have

\begin{equation}
\int_\Gamma \frac{zdz}{z^2-4z+5}
\end{equation}

Let
\begin{equation}
f(z) = \frac{z}{z^2-4z+5} = \frac{z}{(z-(2-i))(z-(2+i))}
\end{equation}

Question a:
The point $2-i$ is outside of the contour $\Gamma$ and the point $2+i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{z}{z-2+i} \\
g(2+i) = \frac{2+i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2+i))}dz = 2\pi i g(z_0) = 2\pi i g(2+i) = 2\pi i \cdot \frac{2+i}{2i}= \pi(2+i)
\end{equation}

Question b:
The point $2+i$ is outside of the contour $\Gamma$ and the point $2-i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{z}{z-2-i} \\
g(2-i) = -\frac{2-i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2-i))}dz = 2\pi i g(z_0) = 2\pi i g(2-i) = 2\pi i \cdot -\frac{2-i}{2i}= -\pi(2-i)
\end{equation}

Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have
\begin{equation}
z_0 = 2+i \\
z_1 = 2-i \\
\left.Res(f;2+i) = \frac{z}{z-2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2-i) = \frac{z}{z-2-i} \right\vert_{z=2-i} = - \frac{2-i}{2i}
\end{equation}
So the Residue Theorem gives us
\begin{equation}
\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i
\end{equation}
« Last Edit: November 24, 2018, 06:22:00 AM by ZhenDi Pan »

#### Yifei Wang

• Jr. Member
•  • Posts: 10
• Karma: 3 ##### Re: TT2A Problem 1
« Reply #2 on: November 24, 2018, 05:34:25 AM »
We can rewrite the fraction as:

$let f(z) =\frac{z}{z^2-4z+5}$

as

$\frac{z}{(z-(2+i))(z-(2-i))}$

a. When $2+i$ is inside
$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2+i}}{z-2-i}~dz= 2i\pi *f(2+i) = \pi * (2+i)$

b. When $2-i$ is inside

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz$

By Cauchy's thm

$f(z) = \int \frac{\frac{z}{z-2-i}}{z-2+i}~dz= 2i\pi *f(2-i) = -\pi * (2-i)$

c. When both points are inside

$f(z) = \int \frac{z}{(z-(2+i))(z-(2-i))}~dz = 2i\pi (Res(f, 2+i) + Res(f, 2-i)) = 2\pi i$

« Last Edit: November 25, 2018, 02:38:57 PM by Yifei Wang »

#### Yifei Wang

• Jr. Member
•  • Posts: 10
• Karma: 3 ##### Re: TT2A Problem 1
« Reply #3 on: November 24, 2018, 05:36:15 AM »

ZhenDi Pan

I think you are missing the $z$ on the numerator.

#### ZhenDi Pan

• Jr. Member
•  • Posts: 10
• Karma: 20 ##### Re: TT2A Problem 1
« Reply #4 on: November 24, 2018, 06:23:00 AM »
Yes thank you I corrected it. Still our answers are different, I don't know where went wrong though.

#### Zhuoer Sun

• Newbie
• • Posts: 1
• Karma: 1 ##### Re: TT2A Problem 1
« Reply #5 on: November 24, 2018, 12:50:28 PM »
Yifei Wang

I think you did the last part wrong. For part (c), you don't need to multiply by 2ipi again. The answer should be the sum of what you got from part (a) and (b), as you've already included 2ipi in previous parts. The final answer should just be pi∗(2+i)−pi∗(2−i).

#### Yifei Wang

• Jr. Member
•  • Posts: 10
• Karma: 3 ##### Re: TT2A Problem 1
« Reply #6 on: November 25, 2018, 02:38:21 PM »
Thank you for the correction!

#### Victor Ivrii ##### Re: TT2A Problem 1
« Reply #7 on: November 29, 2018, 07:57:21 AM »
Remark: Since integrand is $\frac{1}{z} +O(\frac{1}{z^2}$ the residue at $\infty$ is $-1$ and answer to (c) is $2\pi i$ (independently from(b),(c)). Yet another solution to (c)