Evaluate the integral ∮

𝑧−23

𝑧2−4𝑧−5

𝑑𝑧⁡

𝐶

in couterclockwise direction where 𝐶: |𝑧 − 2| = 4 using

Cauchy’s Residue theorem.

To evaluate the integral \(\oint_C \frac{z - 2}{z^2 - 4z - 5} \, dz\) in the counterclockwise direction around the contour \(C: |z - 2| = 4\) using Cauchy's Residue Theorem, we follow these steps:

### Step 1: Identify the poles of the integrand
The integrand is \(\frac{z - 2}{z^2 - 4z - 5}\). We need to factor the denominator to find the poles:
\[
z^2 - 4z - 5 = (z - 5)(z + 1).
\]
So, the integrand has poles at \(z = 5\) and \(z = -1\).

### Step 2: Determine if the poles are inside the contour
The contour \(C\) is the circle \(|z - 2| = 4\). This is a circle centered at \(z = 2\) with a radius of 4. We need to check whether the poles \(z = 5\) and \(z = -1\) lie inside this circle.

- The distance from the center \(z = 2\) to the pole \(z = 5\) is:
  \[
  |5 - 2| = 3.
  \]
  Since \(3 < 4\), the pole \(z = 5\) is inside the contour.
  
- The distance from the center \(z = 2\) to the pole \(z = -1\) is:
  \[
  |-1 - 2| = 3.
  \]
  Since \(3 < 4\), the pole \(z = -1\) is also inside the contour.

Both poles \(z = 5\) and \(z = -1\) are inside the contour \(C\).

### Step 3: Apply Cauchy's Residue Theorem
Cauchy's Residue Theorem states that if a function \(f(z)\) is analytic inside and on a simple closed contour \(C\), except for isolated singularities (poles), then the integral of \(f(z)\) around \(C\) is given by:
\[
\oint_C f(z) \, dz = 2\pi i \sum \text{Residues of } f(z) \text{ inside } C.
\]
We need to compute the residues of the integrand \(\frac{z - 2}{(z - 5)(z + 1)}\) at the poles \(z = 5\) and \(z = -1\).

### Step 4: Calculate the residues
1. **Residue at \(z = 5\):**
   The residue at \(z = 5\) can be found as:
   \[
   \text{Residue}\left(\frac{z - 2}{(z - 5)(z + 1)}, z = 5\right) = \lim_{z \to 5} (z - 5) \frac{z - 2}{(z - 5)(z + 1)} = \frac{5 - 2}{5 + 1} = \frac{3}{6} = \frac{1}{2}.
   \]

2. **Residue at \(z = -1\):**
   The residue at \(z = -1\) is calculated similarly:
   \[
   \text{Residue}\left(\frac{z - 2}{(z - 5)(z + 1)}, z = -1\right) = \lim_{z \to -1} (z + 1) \frac{z - 2}{(z - 5)(z + 1)} = \frac{-1 - 2}{-1 - 5} = \frac{-3}{-6} = \frac{1}{2}.
   \]

### Step 5: Compute the integral
According to Cauchy's Residue Theorem, the value of the integral is:
\[
\oint_C \frac{z - 2}{z^2 - 4z - 5} \, dz = 2\pi i \left( \frac{1}{2} + \frac{1}{2} \right) = 2\pi i \cdot 1 = 2\pi i.
\]

### Final Answer:
The value of the contour integral is:
\[
\boxed{2\pi i}.
\]

Do you want more details, or do you have any questions?

---

**Further questions:**
1. How does the location of poles affect the value of the integral?
2. What is the significance of the Residue Theorem in complex analysis?
3. How would the result change if the contour did not enclose one or both poles?
4. What happens if the contour encloses a higher-order pole?
5. Can this method be extended to integrals over contours in other complex shapes?

**Tip:** When using the Residue Theorem, always ensure that all poles inside the contour are correctly identified and their residues are accurately calculated.

Learn how to evaluate the contour integral ∮ \frac{z - 2}{z^2 - 4z - 5} \, dz \text{ in counterclockwise direction around the circle } C: |z - 2| = 4 \text{ using Cauchy’s Residue theorem. Understand the application of Cauchy's Residue Theorem in complex analysis. Suitable for college-level students.

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