TT2A Problem 1

[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.

[Zhuoer Sun]

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]

Thank you for the correction!

[Victor Ivrii]

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)