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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 β«πΎπ§Β―ππ§ for where πΎ is the circle |π§βπ|=3 oriented counterclockwise I get the following: β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π$π+3πππ‘$=β«2π03π$βπ+3πβππ‘$πππ‘ππ‘=18ππ If I directly apply the Residue Theorem, I would get β«πΎπ§Β―ππ§=2ππRes$π,π§=0$=2ππ

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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 β«πΎπ§Β―ππ§ for where πΎ is the circle |π§βπ|=3 oriented counterclockwise I get the following: β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π$π+3πππ‘$=β«2π03π$βπ+3πβππ‘$πππ‘ππ‘=18ππ If I directly apply the Residue Theorem, I would get β«πΎπ§Β―ππ§=2ππRes$π,π§=0$=2ππ

Miles
4.9
To find the contour integral β«πΎπ§Β―ππ§ over the circle |π§βπ|=3 oriented counterclockwise, we can use two approaches: direct evaluation and using the Residue Theorem.

Approach 1: Direct Evaluation

Let's parameterize the circle πΎ as π§=π+3πππ‘ for π‘ in the range [0, 2π]. We have ππ§=3πππ‘ππ‘.

Substituting these values into the integral, we get:

β«2π0π+3πππ‘β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―π$π+3πππ‘$=β«2π03π$βπ+3πβππ‘$πππ‘ππ‘.

Expanding the expression inside the integral, we have:

β«2π03π$βππππ‘+3πβππ‘πππ‘$ππ‘

Simplifying further, we get:

β«2π03π$βππππ‘+3π0$ππ‘

Now, integrating each term, we have:

β«2π03π$βππππ‘+3$ππ‘ = $βππππ‘+3π‘$|2π0

Plugging in the limits, we get:

$βπππ(2π$+3$2π$)β$βπππ(0$+3$0$)

Simplifying, we have:

$βπ(1$+6π)β$βπ(1$+0) = 6ππ

Therefore, the value of the contour integral is 6ππ.

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 2ππ 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§β0 π§Β― = 1.

Therefore, according to the Residue Theorem, the contour integral β«πΎπ§Β―ππ§ is equal to 2ππ times the residue:

β«πΎπ§Β―ππ§ = 2ππRes$π,π§=0$ = 2ππ$1$ = 2ππ.

Answer: The value of β«πΎπ§Β―ππ§ over the circle |π§βπ|=3 oriented counterclockwise is 2ππ.

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