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 ∫𝛾𝑧¯𝑑𝑧 for where 𝛾 is the circle |𝑧−𝑖|=3 oriented counterclockwise I get the following: ∫2𝜋0𝑖+3𝑒𝑖𝑡⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑(𝑖+3𝑒𝑖𝑡)=∫2𝜋03𝑖(−𝑖+3𝑒−𝑖𝑡)𝑒𝑖𝑡𝑑𝑡=18𝜋𝑖 If I directly apply the Residue Theorem, I would get ∫𝛾𝑧¯𝑑𝑧=2𝜋𝑖Res(𝑓,𝑧=0)=2𝜋𝑖

To find the contour integral ∫𝛾𝑧¯𝑑𝑧 over the circle |𝑧−𝑖|=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+3𝑒𝑖𝑡 for 𝑡 in the range [0, 2𝜋]. We have 𝑑𝑧=3𝑒𝑖𝑡𝑑𝑡.

Substituting these values into the integral, we get:

∫2𝜋0𝑖+3𝑒𝑖𝑡⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯𝑑(𝑖+3𝑒𝑖𝑡)=∫2𝜋03𝑖(−𝑖+3𝑒−𝑖𝑡)𝑒𝑖𝑡𝑑𝑡.

Expanding the expression inside the integral, we have:

∫2𝜋03𝑖(−𝑖𝑒𝑖𝑡+3𝑒−𝑖𝑡𝑒𝑖𝑡)𝑑𝑡

Simplifying further, we get:

∫2𝜋03𝑖(−𝑖𝑒𝑖𝑡+3𝑒0)𝑑𝑡

Now, integrating each term, we have:

∫2𝜋03𝑖(−𝑖𝑒𝑖𝑡+3)𝑑𝑡 = (−𝑖𝑒𝑖𝑡+3𝑡)|2𝜋0

Plugging in the limits, we get:

(−𝑖𝑒𝑖(2𝜋)+3(2𝜋))−(−𝑖𝑒𝑖(0)+3(0))

Simplifying, we have:

(−𝑖(1)+6𝜋)−(−𝑖(1)+0) = 6𝜋𝑖

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

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 2𝜋𝑖 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 ∫𝛾𝑧¯𝑑𝑧 is equal to 2𝜋𝑖 times the residue:

∫𝛾𝑧¯𝑑𝑧 = 2𝜋𝑖Res(𝑓,𝑧=0) = 2𝜋𝑖(1) = 2𝜋𝑖.

Answer: The value of ∫𝛾𝑧¯𝑑𝑧 over the circle |𝑧−𝑖|=3 oriented counterclockwise is 2𝜋𝑖.