### Author Topic: TT2 Problem 1  (Read 1677 times)

#### Victor Ivrii

• Elder Member
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##### TT2 Problem 1
« on: November 24, 2018, 04:27:17 AM »
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

• Newbie
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##### Re: TT2 Problem 1
« Reply #1 on: November 24, 2018, 04:46:09 AM »
this is my solution

#### ZhenDi Pan

• Jr. Member
• Posts: 10
• Karma: 20
##### Re: TT2 Problem 1
« Reply #2 on: November 24, 2018, 04:49:12 AM »
We have

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

Let

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

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

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}

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

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}

Question c:
Both points $1+3i$ and $1-3i$ are inside of the coutour $\Gamma$. Then we have

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}

So the Residue Theorem gives us

\int_\Gamma f(z)dz = 2\pi i(-\frac{1}{6i}+\frac{1}{6i}) = 2\pi i \cdot 0 = 0
« Last Edit: November 24, 2018, 05:43:26 AM by ZhenDi Pan »

#### 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)