Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:54:06 AM

Using Cauchy's integral formula calculate
$$
\int_\Gamma \frac{z\,dz}{z^24z+5},
$$
where $\Gamma$ is a counterclockwise oriented simple contour, not passing through eiter
of $2\pm i$ in the following cases
(a) The point $2+i$ is inside $\Gamma$ and $2i$ is outside it;
(b) The point $2i$ is inside $\Gamma$ and $2+i$ is outside it;
(c) Both points $2\pm i$ are inside $\Gamma$.

We have
\begin{equation}
\int_\Gamma \frac{zdz}{z^24z+5}
\end{equation}
Let
\begin{equation}
f(z) = \frac{z}{z^24z+5} = \frac{z}{(z(2i))(z(2+i))}
\end{equation}
Question a:
The point $2i$ 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}{z2+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 $2i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{z}{z2i} \\
g(2i) = \frac{2i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z(2i))}dz = 2\pi i g(z_0) = 2\pi i g(2i) = 2\pi i \cdot \frac{2i}{2i}= \pi(2i)
\end{equation}
Question c:
Both points $2+i$ and $2i$ are inside of the coutour $\Gamma$. Then we have
\begin{equation}
z_0 = 2+i \\
z_1 = 2i \\
\left.Res(f;2+i) = \frac{z}{z2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2i) = \frac{z}{z2i} \right\vert_{z=2i} =  \frac{2i}{2i}
\end{equation}
So the Residue Theorem gives us
\begin{equation}
\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}\frac{2i}{2i}) = 2\pi i \cdot 1= 2\pi i
\end{equation}

We can rewrite the fraction as:
$let f(z) =\frac{z}{z^24z+5}$
as
$\frac{z}{(z(2+i))(z(2i))}$
a. When $2+i$ is inside
$f(z) = \int \frac{\frac{z}{z2+i}}{z2i}~dz$
By Cauchy's thm
$f(z) = \int \frac{\frac{z}{z2+i}}{z2i}~dz= 2i\pi *f(2+i) = \pi * (2+i)$
b. When $2i$ is inside
$f(z) = \int \frac{\frac{z}{z2i}}{z2+i}~dz$
By Cauchy's thm
$f(z) = \int \frac{\frac{z}{z2i}}{z2+i}~dz= 2i\pi *f(2i) = \pi * (2i)$
c. When both points are inside
$f(z) = \int \frac{z}{(z(2+i))(z(2i))}~dz = 2i\pi (Res(f, 2+i) + Res(f, 2i)) = 2\pi i$

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

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

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).

Thank you for the correction!

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)