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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 βˆ«π›Ύπ‘§Β―π‘‘π‘§ for where 𝛾 is the circle |π‘§βˆ’π‘–|=3 oriented counterclockwise I get the following: ∫2πœ‹0𝑖+3π‘’π‘–π‘‘βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―π‘‘(𝑖+3𝑒𝑖𝑑)=∫2πœ‹03𝑖(βˆ’π‘–+3π‘’βˆ’π‘–π‘‘)𝑒𝑖𝑑𝑑𝑑=18πœ‹π‘– If I directly apply the Residue Theorem, I would get βˆ«π›Ύπ‘§Β―π‘‘π‘§=2πœ‹π‘–Res(𝑓,𝑧=0)=2πœ‹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 βˆ«π›Ύπ‘§Β―π‘‘π‘§ for where 𝛾 is the circle |π‘§βˆ’π‘–|=3 oriented counterclockwise I get the following: ∫2πœ‹0𝑖+3π‘’π‘–π‘‘βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―π‘‘(𝑖+3𝑒𝑖𝑑)=∫2πœ‹03𝑖(βˆ’π‘–+3π‘’βˆ’π‘–π‘‘)𝑒𝑖𝑑𝑑𝑑=18πœ‹π‘– If I directly apply the Residue Theorem, I would get βˆ«π›Ύπ‘§Β―π‘‘π‘§=2πœ‹π‘–Res(𝑓,𝑧=0)=2πœ‹π‘–

Expert avatar
Miles
4.9
110 Answers
To find the contour integral βˆ«π›Ύπ‘§Β―π‘‘π‘§ over the circle |π‘§βˆ’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 𝛾 as 𝑧=𝑖+3𝑒𝑖𝑑 for 𝑑 in the range [0, 2πœ‹]. We have 𝑑𝑧=3𝑒𝑖𝑑𝑑𝑑.

Substituting these values into the integral, we get:

∫2πœ‹0𝑖+3π‘’π‘–π‘‘βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―βŽ―π‘‘(𝑖+3𝑒𝑖𝑑)=∫2πœ‹03𝑖(βˆ’π‘–+3π‘’βˆ’π‘–π‘‘)𝑒𝑖𝑑𝑑𝑑.

Expanding the expression inside the integral, we have:

∫2πœ‹03𝑖(βˆ’π‘–π‘’π‘–π‘‘+3π‘’βˆ’π‘–π‘‘π‘’π‘–π‘‘)𝑑𝑑

Simplifying further, we get:

∫2πœ‹03𝑖(βˆ’π‘–π‘’π‘–π‘‘+3𝑒0)𝑑𝑑

Now, integrating each term, we have:

∫2πœ‹03𝑖(βˆ’π‘–π‘’π‘–π‘‘+3)𝑑𝑑 = (βˆ’π‘–π‘’π‘–π‘‘+3𝑑)|2πœ‹0

Plugging in the limits, we get:

(βˆ’π‘–π‘’π‘–(2πœ‹)+3(2πœ‹))βˆ’(βˆ’π‘–π‘’π‘–(0)+3(0))

Simplifying, we have:

(βˆ’π‘–(1)+6πœ‹)βˆ’(βˆ’π‘–(1)+0) = 6πœ‹π‘–

Therefore, the value of the contour integral is 6πœ‹π‘–.

Approach 2: Using the Residue Theorem

According to the Residue Theorem, the contour integral of a function 𝑓(𝑧) over a closed curve is equal to 2πœ‹π‘– times the sum of residues of 𝑓(𝑧) at its singularities inside the curve.

In this case, we need to find the residue of the function 𝑓(𝑧) = 𝑧¯ at the singularity 𝑧=0.

The residue at 𝑧=0 can be found by taking the limit of 𝑧¯ as 𝑧 approaches 0:

Res(𝑓,𝑧=0) = lim𝑧→0 𝑧¯ = 1.

Therefore, according to the Residue Theorem, the contour integral βˆ«π›Ύπ‘§Β―π‘‘π‘§ is equal to 2πœ‹π‘– times the residue:

βˆ«π›Ύπ‘§Β―π‘‘π‘§ = 2πœ‹π‘–Res(𝑓,𝑧=0) = 2πœ‹π‘–(1) = 2πœ‹π‘–.

Answer: The value of βˆ«π›Ύπ‘§Β―π‘‘π‘§ over the circle |π‘§βˆ’π‘–|=3 oriented counterclockwise is 2πœ‹π‘–.

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