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 β«πΎπ§Β―ππ§ for where πΎ is the circle |π§βπ|=3 oriented counterclockwise I get the following: β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π(π+3πππ‘)=β«2π03π(βπ+3πβππ‘)πππ‘ππ‘=18ππ If I directly apply the Residue Theorem, I would get β«πΎπ§Β―ππ§=2ππRes(π,π§=0)=2ππ

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Miles

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To find the contour integral β«πΎπ§Β―ππ§ over the circle |π§βπ|=3 oriented counterclockwise, we can use two approaches: direct evaluation and using the Residue Theorem.

Approach 1: Direct Evaluation

Let's parameterize the circle πΎ as π§=π+3πππ‘ for π‘ in the range [0, 2π]. We have ππ§=3πππ‘ππ‘.

Substituting these values into the integral, we get:

β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π(π+3πππ‘)=β«2π03π(βπ+3πβππ‘)πππ‘ππ‘.

Expanding the expression inside the integral, we have:

β«2π03π(βππππ‘+3πβππ‘πππ‘)ππ‘

Simplifying further, we get:

β«2π03π(βππππ‘+3π0)ππ‘

Now, integrating each term, we have:

β«2π03π(βππππ‘+3)ππ‘ = (βππππ‘+3π‘)|2π0

Plugging in the limits, we get:

(βπππ(2π)+3(2π))β(βπππ(0)+3(0))

Simplifying, we have:

(βπ(1)+6π)β(βπ(1)+0) = 6ππ

Therefore, the value of the contour integral is 6ππ.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function π(π§) over a closed curve is equal to 2ππ times the sum of residues of π(π§) at its singularities inside the curve.

In this case, we need to find the residue of the function π(π§) = π§Β― at the singularity π§=0.

The residue at π§=0 can be found by taking the limit of π§Β― as π§ approaches 0:

Res(π,π§=0) = limπ§β0 π§Β― = 1.

Therefore, according to the Residue Theorem, the contour integral β«πΎπ§Β―ππ§ is equal to 2ππ times the residue:

β«πΎπ§Β―ππ§ = 2ππRes(π,π§=0) = 2ππ(1) = 2ππ.

Answer: The value of β«πΎπ§Β―ππ§ over the circle |π§βπ|=3 oriented counterclockwise is 2ππ.

Approach 1: Direct Evaluation

Let's parameterize the circle πΎ as π§=π+3πππ‘ for π‘ in the range [0, 2π]. We have ππ§=3πππ‘ππ‘.

Substituting these values into the integral, we get:

β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π(π+3πππ‘)=β«2π03π(βπ+3πβππ‘)πππ‘ππ‘.

Expanding the expression inside the integral, we have:

β«2π03π(βππππ‘+3πβππ‘πππ‘)ππ‘

Simplifying further, we get:

β«2π03π(βππππ‘+3π0)ππ‘

Now, integrating each term, we have:

β«2π03π(βππππ‘+3)ππ‘ = (βππππ‘+3π‘)|2π0

Plugging in the limits, we get:

(βπππ(2π)+3(2π))β(βπππ(0)+3(0))

Simplifying, we have:

(βπ(1)+6π)β(βπ(1)+0) = 6ππ

Therefore, the value of the contour integral is 6ππ.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function π(π§) over a closed curve is equal to 2ππ times the sum of residues of π(π§) at its singularities inside the curve.

In this case, we need to find the residue of the function π(π§) = π§Β― at the singularity π§=0.

The residue at π§=0 can be found by taking the limit of π§Β― as π§ approaches 0:

Res(π,π§=0) = limπ§β0 π§Β― = 1.

Therefore, according to the Residue Theorem, the contour integral β«πΎπ§Β―ππ§ is equal to 2ππ times the residue:

β«πΎπ§Β―ππ§ = 2ππRes(π,π§=0) = 2ππ(1) = 2ππ.

Answer: The value of β«πΎπ§Β―ππ§ over the circle |π§βπ|=3 oriented counterclockwise is 2ππ.

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