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 โซ๐พ๐งยฏ๐๐ง for where ๐พ is the circle |๐งโ๐|=3 oriented counterclockwise I get the following: โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐(๐+3๐๐๐ก)=โซ2๐03๐(โ๐+3๐โ๐๐ก)๐๐๐ก๐๐ก=18๐๐ If I directly apply the Residue Theorem, I would get โซ๐พ๐งยฏ๐๐ง=2๐๐Res(๐,๐ง=0)=2๐๐

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To find the contour integral โซ๐พ๐งยฏ๐๐ง over the circle |๐งโ๐|=3 oriented counterclockwise, we can use two approaches: direct evaluation and using the Residue Theorem.

Approach 1: Direct Evaluation

Let's parameterize the circle ๐พ as ๐ง=๐+3๐๐๐ก for ๐ก in the range [0, 2๐]. We have ๐๐ง=3๐๐๐ก๐๐ก.

Substituting these values into the integral, we get:

โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐(๐+3๐๐๐ก)=โซ2๐03๐(โ๐+3๐โ๐๐ก)๐๐๐ก๐๐ก.

Expanding the expression inside the integral, we have:

โซ2๐03๐(โ๐๐๐๐ก+3๐โ๐๐ก๐๐๐ก)๐๐ก

Simplifying further, we get:

โซ2๐03๐(โ๐๐๐๐ก+3๐0)๐๐ก

Now, integrating each term, we have:

โซ2๐03๐(โ๐๐๐๐ก+3)๐๐ก = (โ๐๐๐๐ก+3๐ก)|2๐0

Plugging in the limits, we get:

(โ๐๐๐(2๐)+3(2๐))โ(โ๐๐๐(0)+3(0))

Simplifying, we have:

(โ๐(1)+6๐)โ(โ๐(1)+0) = 6๐๐

Therefore, the value of the contour integral is 6๐๐.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function ๐(๐ง) over a closed curve is equal to 2๐๐ times the sum of residues of ๐(๐ง) at its singularities inside the curve.

In this case, we need to find the residue of the function ๐(๐ง) = ๐งยฏ at the singularity ๐ง=0.

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

Res(๐,๐ง=0) = lim๐งโ0 ๐งยฏ = 1.

Therefore, according to the Residue Theorem, the contour integral โซ๐พ๐งยฏ๐๐ง is equal to 2๐๐ times the residue:

โซ๐พ๐งยฏ๐๐ง = 2๐๐Res(๐,๐ง=0) = 2๐๐(1) = 2๐๐.

Answer: The value of โซ๐พ๐งยฏ๐๐ง over the circle |๐งโ๐|=3 oriented counterclockwise is 2๐๐.

Approach 1: Direct Evaluation

Let's parameterize the circle ๐พ as ๐ง=๐+3๐๐๐ก for ๐ก in the range [0, 2๐]. We have ๐๐ง=3๐๐๐ก๐๐ก.

Substituting these values into the integral, we get:

โซ2๐0๐+3๐๐๐กโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏโฏ๐(๐+3๐๐๐ก)=โซ2๐03๐(โ๐+3๐โ๐๐ก)๐๐๐ก๐๐ก.

Expanding the expression inside the integral, we have:

โซ2๐03๐(โ๐๐๐๐ก+3๐โ๐๐ก๐๐๐ก)๐๐ก

Simplifying further, we get:

โซ2๐03๐(โ๐๐๐๐ก+3๐0)๐๐ก

Now, integrating each term, we have:

โซ2๐03๐(โ๐๐๐๐ก+3)๐๐ก = (โ๐๐๐๐ก+3๐ก)|2๐0

Plugging in the limits, we get:

(โ๐๐๐(2๐)+3(2๐))โ(โ๐๐๐(0)+3(0))

Simplifying, we have:

(โ๐(1)+6๐)โ(โ๐(1)+0) = 6๐๐

Therefore, the value of the contour integral is 6๐๐.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function ๐(๐ง) over a closed curve is equal to 2๐๐ times the sum of residues of ๐(๐ง) at its singularities inside the curve.

In this case, we need to find the residue of the function ๐(๐ง) = ๐งยฏ at the singularity ๐ง=0.

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

Res(๐,๐ง=0) = lim๐งโ0 ๐งยฏ = 1.

Therefore, according to the Residue Theorem, the contour integral โซ๐พ๐งยฏ๐๐ง is equal to 2๐๐ times the residue:

โซ๐พ๐งยฏ๐๐ง = 2๐๐Res(๐,๐ง=0) = 2๐๐(1) = 2๐๐.

Answer: The value of โซ๐พ๐งยฏ๐๐ง over the circle |๐งโ๐|=3 oriented counterclockwise is 2๐๐.

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