# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:54:06 AM

Title: TT2A Problem 1
Post by: Victor Ivrii 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  $\Gamma$.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan on November 24, 2018, 05:30:25 AM
We have

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

Let

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

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

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)

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

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)

Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have

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}

So the Residue Theorem gives us

\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i
Title: Re: TT2A Problem 1
Post by: Yifei Wang 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

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

Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 24, 2018, 05:36:15 AM

ZhenDi Pan

I think you are missing the $z$ on the numerator.
Title: Re: TT2A Problem 1
Post by: ZhenDi Pan 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.
Title: Re: TT2A Problem 1
Post by: Zhuoer Sun 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).

Title: Re: TT2A Problem 1
Post by: Yifei Wang on November 25, 2018, 02:38:21 PM
Thank you for the correction!
Title: Re: TT2A Problem 1
Post by: Victor Ivrii 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)