### Author Topic: TT2B Problem 1  (Read 5156 times)

#### Victor Ivrii

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##### TT2B Problem 1
« on: November 24, 2018, 05:19:15 AM »
Using Cauchy's integral formula calculate
$$\int_\Gamma \frac{dz}{z^2-6z+25},$$
where $\Gamma$ is a counter-clockwise oriented simple contour, not passing through eiter
of $1\pm 3i$ in the following cases

(a) The point $3+4i$ is inside  $\Gamma$ and $3-4i$ is outside  it;

(b) The point $3-4i$ is inside  $\Gamma$ and $3+4i$ is outside it;

(c) Both points $3\pm 4i$ are inside  $\Gamma$.

#### Sojung Lee

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##### Re: TT2B Problem 1
« Reply #1 on: November 24, 2018, 06:17:50 AM »
Cauchy's integral formula is $$f(z) = \frac{1}{2𝜋i} \int_{Γ}\frac{f(z)}{z-a} dz$$
given formula is $$\int_{Γ}\frac{dz}{z^2-6z+25} = \int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$
For (a), as 3+4i is inside and 3-4i is outside
Let $$f(z)=\frac{1}{z-(3-4i)}$$
by Cauchy theorem $$\int_{Γ}\frac{f(z) dz}{(z-(3+4i))}=2𝜋i f(3+4i) = \frac{2𝜋i}{8i} = \frac{𝜋}{4}$$
For (b), as 3-4i is inside and 3+4i is outside
Let $$f(z)=\frac{1}{z-(3+4i)}$$
by Cauchy theorem $$\int_{Γ}\frac{f(z) dz}{(z-(3-4i))}=2𝜋i f(3-4i) = \frac{2𝜋i}{-8i} = \frac{-𝜋}{4}$$
For (c), as both points 3±4i are inside, we can use residue theorem
$$\int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$
$$=2𝜋i(Res(f, 3+-4i) + Res(f, 3+4i))$$
$$=2𝜋i(\frac{1}{8i}+\frac{-1}{8i})=0$$
« Last Edit: November 24, 2018, 10:37:15 PM by Sojung Lee »

#### hanyu Qi

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##### Re: TT2B Problem 1
« Reply #2 on: November 24, 2018, 05:37:34 PM »
just want to add a little detail on (c), if both points are inside we can use residue theorem so we can get 2πI ( res(f,3+4i)+res(f,3-4i)).

#### Sojung Lee

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##### Re: TT2B Problem 1
« Reply #3 on: November 24, 2018, 10:38:02 PM »
Yes, I added your comment. Thanks!
« Last Edit: November 24, 2018, 11:50:52 PM by Sojung Lee »