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# I have a complex function I would like to integrate over. I can use two approaches and they should give the same solution. If I want to find the contour integral โซ๐พ๐งยฏ๐๐ง for where ๐พ is the circle |๐งโ๐|=3 oriented counterclockwise I get the following: โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐$๐+3๐๐๐ก$=โซ2๐03๐$โ๐+3๐โ๐๐ก$๐๐๐ก๐๐ก=18๐๐ If I directly apply the Residue Theorem, I would get โซ๐พ๐งยฏ๐๐ง=2๐๐Res$๐,๐ง=0$=2๐๐

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## Answer to a math question I have a complex function I would like to integrate over. I can use two approaches and they should give the same solution. If I want to find the contour integral โซ๐พ๐งยฏ๐๐ง for where ๐พ is the circle |๐งโ๐|=3 oriented counterclockwise I get the following: โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐$๐+3๐๐๐ก$=โซ2๐03๐$โ๐+3๐โ๐๐ก$๐๐๐ก๐๐ก=18๐๐ If I directly apply the Residue Theorem, I would get โซ๐พ๐งยฏ๐๐ง=2๐๐Res$๐,๐ง=0$=2๐๐

Miles
4.9
To find the contour integral โซ๐พ๐งยฏ๐๐ง over the circle |๐งโ๐|=3 oriented counterclockwise, we can use two approaches: direct evaluation and using the Residue Theorem.

Approach 1: Direct Evaluation

Let's parameterize the circle ๐พ as ๐ง=๐+3๐๐๐ก for ๐ก in the range [0, 2๐]. We have ๐๐ง=3๐๐๐ก๐๐ก.

Substituting these values into the integral, we get:

โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐$๐+3๐๐๐ก$=โซ2๐03๐$โ๐+3๐โ๐๐ก$๐๐๐ก๐๐ก.

Expanding the expression inside the integral, we have:

โซ2๐03๐$โ๐๐๐๐ก+3๐โ๐๐ก๐๐๐ก$๐๐ก

Simplifying further, we get:

โซ2๐03๐$โ๐๐๐๐ก+3๐0$๐๐ก

Now, integrating each term, we have:

โซ2๐03๐$โ๐๐๐๐ก+3$๐๐ก = $โ๐๐๐๐ก+3๐ก$|2๐0

Plugging in the limits, we get:

$โ๐๐๐(2๐$+3$2๐$)โ$โ๐๐๐(0$+3$0$)

Simplifying, we have:

$โ๐(1$+6๐)โ$โ๐(1$+0) = 6๐๐

Therefore, the value of the contour integral is 6๐๐.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function ๐$๐ง$ over a closed curve is equal to 2๐๐ times the sum of residues of ๐$๐ง$ at its singularities inside the curve.

In this case, we need to find the residue of the function ๐$๐ง$ = ๐งยฏ at the singularity ๐ง=0.

The residue at ๐ง=0 can be found by taking the limit of ๐งยฏ as ๐ง approaches 0:

Res$๐,๐ง=0$ = lim๐งโ0 ๐งยฏ = 1.

Therefore, according to the Residue Theorem, the contour integral โซ๐พ๐งยฏ๐๐ง is equal to 2๐๐ times the residue:

โซ๐พ๐งยฏ๐๐ง = 2๐๐Res$๐,๐ง=0$ = 2๐๐$1$ = 2๐๐.

Answer: The value of โซ๐พ๐งยฏ๐๐ง over the circle |๐งโ๐|=3 oriented counterclockwise is 2๐๐.

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