Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 2 => Topic started by: Victor Ivrii on November 24, 2018, 04:27:17 AM

Using Cauchy's integral formula calculate
$$
\int_\Gamma \frac{dz}{z^22z+10},
$$
where $\Gamma$ is a counterclockwise oriented simple contour, not passing through eiter
of $1\pm 3i$ in the following cases
(a) The point $1+3i$ is inside $\Gamma$ and $13i$ is outside it;
(b) The point $13i$ is inside $\Gamma$ and $1+3i$ is outside it;
(c) Both points $1\pm 3i$ are inside $\Gamma$.

this is my solution

We have
\begin{equation}
\int_\Gamma \frac{dz}{z^22z+10}
\end{equation}
Let
\begin{equation}
f(z) = \frac{1}{z^22z+10} = \frac{1}{(z(13i))(z(1+3i))}
\end{equation}
Question a:
The point $13i$ is outside of the contour $\Gamma$ and the point $1+3i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{1}{z1+3i} \\
g(1+3i) = \frac{1}{6i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z(1+3i))}dz = 2\pi i g(z_0) = 2\pi i g(1+3i) = 2\pi i \cdot \frac{1}{6i}=\frac{\pi}{3}
\end{equation}
Question b:
The point $1+3i$ is outside of the contour $\Gamma$ and the point $13i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{1}{z13i} \\
g(13i) = \frac{1}{6i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z(13i))}dz = 2\pi i g(z_0) = 2\pi i g(13i) = 2\pi i \cdot \frac{1}{6i}= \frac{\pi}{3}
\end{equation}
Question c:
Both points $1+3i$ and $13i$ are inside of the coutour $\Gamma$. Then we have
\begin{equation}
z_0 = 1+3i \\
z_1 = 13i \\
\left.Res(f;1+3i) = \frac{1}{z1+3i} \right\vert_{z=1+3i} = \frac{1}{6i} \\
\left.Res(f;1+3i) = \frac{1}{z13i} \right\vert_{z=13i} =  \frac{1}{6i}
\end{equation}
So the Residue Theorem gives us
\begin{equation}
\int_\Gamma f(z)dz = 2\pi i(\frac{1}{6i}+\frac{1}{6i}) = 2\pi i \cdot 0 = 0
\end{equation}

Remark: there is also $\infty$ point (outside of contours) but since the integrand decays as $z^{2}$ at infinity, its residue at $\infty$ is $0$ and therefore answer to (c) is $0$ (yet another proof)