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 โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง for where ๐›พ is the circle |๐‘งโˆ’๐‘–|=3 oriented counterclockwise I get the following: โˆซ2๐œ‹0๐‘–+3๐‘’๐‘–๐‘กโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏ๐‘‘(๐‘–+3๐‘’๐‘–๐‘ก)=โˆซ2๐œ‹03๐‘–(โˆ’๐‘–+3๐‘’โˆ’๐‘–๐‘ก)๐‘’๐‘–๐‘ก๐‘‘๐‘ก=18๐œ‹๐‘– If I directly apply the Residue Theorem, I would get โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง=2๐œ‹๐‘–Res(๐‘“,๐‘ง=0)=2๐œ‹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 โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง for where ๐›พ is the circle |๐‘งโˆ’๐‘–|=3 oriented counterclockwise I get the following: โˆซ2๐œ‹0๐‘–+3๐‘’๐‘–๐‘กโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏ๐‘‘(๐‘–+3๐‘’๐‘–๐‘ก)=โˆซ2๐œ‹03๐‘–(โˆ’๐‘–+3๐‘’โˆ’๐‘–๐‘ก)๐‘’๐‘–๐‘ก๐‘‘๐‘ก=18๐œ‹๐‘– If I directly apply the Residue Theorem, I would get โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง=2๐œ‹๐‘–Res(๐‘“,๐‘ง=0)=2๐œ‹๐‘–

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Miles
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50 Answers
To find the contour integral โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง over the circle |๐‘งโˆ’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œ‹]. We have ๐‘‘๐‘ง=3๐‘’๐‘–๐‘ก๐‘‘๐‘ก.

Substituting these values into the integral, we get:

โˆซ2๐œ‹0๐‘–+3๐‘’๐‘–๐‘กโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏโŽฏ๐‘‘(๐‘–+3๐‘’๐‘–๐‘ก)=โˆซ2๐œ‹03๐‘–(โˆ’๐‘–+3๐‘’โˆ’๐‘–๐‘ก)๐‘’๐‘–๐‘ก๐‘‘๐‘ก.

Expanding the expression inside the integral, we have:

โˆซ2๐œ‹03๐‘–(โˆ’๐‘–๐‘’๐‘–๐‘ก+3๐‘’โˆ’๐‘–๐‘ก๐‘’๐‘–๐‘ก)๐‘‘๐‘ก

Simplifying further, we get:

โˆซ2๐œ‹03๐‘–(โˆ’๐‘–๐‘’๐‘–๐‘ก+3๐‘’0)๐‘‘๐‘ก

Now, integrating each term, we have:

โˆซ2๐œ‹03๐‘–(โˆ’๐‘–๐‘’๐‘–๐‘ก+3)๐‘‘๐‘ก = (โˆ’๐‘–๐‘’๐‘–๐‘ก+3๐‘ก)|2๐œ‹0

Plugging in the limits, we get:

(โˆ’๐‘–๐‘’๐‘–(2๐œ‹)+3(2๐œ‹))โˆ’(โˆ’๐‘–๐‘’๐‘–(0)+3(0))

Simplifying, we have:

(โˆ’๐‘–(1)+6๐œ‹)โˆ’(โˆ’๐‘–(1)+0) = 6๐œ‹๐‘–

Therefore, the value of the contour integral is 6๐œ‹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œ‹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 โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง is equal to 2๐œ‹๐‘– times the residue:

โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง = 2๐œ‹๐‘–Res(๐‘“,๐‘ง=0) = 2๐œ‹๐‘–(1) = 2๐œ‹๐‘–.

Answer: The value of โˆซ๐›พ๐‘งยฏ๐‘‘๐‘ง over the circle |๐‘งโˆ’๐‘–|=3 oriented counterclockwise is 2๐œ‹๐‘–.

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