MAT334-2018F > Term Test 2

TT2A Problem 1

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Victor Ivrii:
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:
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}

Yifei Wang:
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

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

Yifei Wang:

ZhenDi Pan

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

ZhenDi Pan:
Yes thank you I corrected it. Still our answers are different, I don't know where went wrong though.

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