To solve this problem using the Cauchy integral formula, let's analyze the given integral:

Problem:

We are tasked with evaluating: $\oint_C \frac{z-1}{(z+1)^2 (z-2)} \, dz$ where $C$ is the circle $|z-i| = 2$.

Step-by-step Solution:

1. Identify the poles within the contour $C$:

   • The poles of the integrand are at: $z = -1 \quad (\text{double pole due to } (z+1)^2), \quad \text{and } z = 2 \quad (\text{simple pole due to } (z-2)).$
   • Since $|z-i| = 2$ defines a circle centered at $z = i$ with radius 2, let's check which poles are inside the contour.
     • For $z = -1$, $|-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$, which is less than 2. Inside the contour.
     • For $z = 2$, $|2 - i| = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}$, which is greater than 2. Outside the contour.

   Therefore, only $z = -1$ contributes to the integral.

2. Focus on the residue at $z = -1$:

   • Since $z = -1$ is a double pole, we use the formula for residues of a function $f(z)$ at a pole of order 2: $\text{Res}\left(f(z), z = -1\right) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d}{dz}\left[(z - z_0)^m f(z)\right], \quad m = 2.$

   Here, $f(z) = \frac{z-1}{(z+1)^2 (z-2)}$, $z_0 = -1$, and $m = 2$.

3. Simplify for the residue:

   • Multiply the integrand by $(z+1)^2$ to simplify around $z = -1$: $\frac{(z - 1)}{(z+1)^2 (z - 2)} \cdot (z+1)^2 = \frac{z-1}{z-2}.$
   • Differentiate $\frac{z-1}{z-2}$ with respect to $z$: $\frac{d}{dz} \left(\frac{z-1}{z-2}\right) = \frac{(z-2)(1) - (z-1)(1)}{(z-2)^2} = \frac{z-2 - z + 1}{(z-2)^2} = \frac{-1}{(z-2)^2}.$
   • Evaluate the derivative at $z = -1$: $\frac{-1}{((-1) - 2)^2} = \frac{-1}{(-3)^2} = \frac{-1}{9}.$

   Therefore, the residue at $z = -1$ is $\frac{-1}{9}$.

4. Apply the Cauchy integral formula:

   • By the Cauchy integral formula, the integral is: $\oint_C f(z) \, dz = 2\pi i \cdot \text{Res}(f(z), z = -1).$
   • Substitute the residue: $\oint_C \frac{z-1}{(z+1)^2 (z-2)} \, dz = 2\pi i \cdot \frac{-1}{9} = \frac{-2\pi i}{9}.$

Final Answer:

$\boxed{\frac{-2\pi i}{9}}$

Let me know if you'd like further clarification or additional examples.

Related Questions:

1. What is the general formula for residues at higher-order poles?
2. How does the choice of contour affect which poles contribute to the integral?
3. What is the Cauchy integral formula for simple poles?
4. How can this problem be extended to include multiple poles within the contour?
5. How would the solution change if $|z - i| = 4$ instead of 2?

Tip:

Always verify which poles lie inside the given contour before proceeding with residue calculations.