TT2 Problem 1

[Victor Ivrii]

Using Cauchy's integral formula calculate
$$
\int_\Gamma \frac{dz}{z^2-2z+10},
$$
where $\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter
of $1\pm 3i$ in the following cases

(a) The point $1+3i$ is inside  $\Gamma$ and $1-3i$ is outside  it;

(b)  The point $1-3i$ is inside  $\Gamma$ and $1+3i$ is outside it;

(c) Both points $1\pm 3i$ are inside  $\Gamma$.

[Hongyue Zhu]

this is my solution

[ZhenDi Pan]

We have

\begin{equation}
\int_\Gamma \frac{dz}{z^2-2z+10} 
\end{equation}

Let
\begin{equation}
f(z) = \frac{1}{z^2-2z+10} = \frac{1}{(z-(1-3i))(z-(1+3i))}
\end{equation}

Question a:
The point $1-3i$ 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}{z-1+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 $1-3i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{1}{z-1-3i} \\
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 c:
Both points $1+3i$ and $1-3i$ are inside of the coutour $\Gamma$. Then we have
\begin{equation}
z_0 = 1+3i \\
z_1 = 1-3i \\
\left.Res(f;1+3i) = \frac{1}{z-1+3i} \right\vert_{z=1+3i} = \frac{1}{6i} \\
\left.Res(f;1+3i) = \frac{1}{z-1-3i} \right\vert_{z=1-3i} = - \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}

[Victor Ivrii]

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)