To solve the given integral using the residue theorem, we start with the integral expression:

$\int_{C} \frac{z \, dz}{(z-1)(z-2)^2}$

where $C$ is the contour $|z| = 3$.

The function to integrate is:

$f(z) = \frac{z}{(z-1)(z-2)^2}$

This function has poles at $z = 1$ and $z = 2$. Since the contour $|z| = 3$ encloses both of these poles, we need to calculate the residues at each pole.

Residue at $z = 1$:

The residue at a simple pole $z = 1$ can be found using the formula:

$\text{Res}(f, 1) = \lim_{z \to 1} (z-1) f(z) = \lim_{z \to 1} \frac{z}{(z-2)^2}$

Evaluating this limit:

$\text{Res}(f, 1) = \frac{1}{(1-2)^2} = \frac{1}{1} = 1$

Residue at $z = 2$:

The residue at a pole of order 2, such as $z = 2$, can be found using the formula:

$\text{Res}(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left[ (z-2)^2 \frac{z}{(z-1)(z-2)^2} \right]$

Simplifying inside the derivative:

$\text{Res}(f, 2) = \lim_{z \to 2} \frac{d}{dz} \left[ \frac{z}{z-1} \right]$

Differentiating:

$\frac{d}{dz} \left[ \frac{z}{z-1} \right] = \frac{(z-1) - z}{(z-1)^2} = \frac{-1}{(z-1)^2}$

Evaluating this derivative at $z = 2$:

$\text{Res}(f, 2) = \frac{-1}{(2-1)^2} = -1$

Sum of residues:

Now, we sum the residues:

$\sum \text{Res}(f, z_i) = 1 + (-1) = 0$

Conclusion:

By the residue theorem, the integral is:

$\int_{C} \frac{z \, dz}{(z-1)(z-2)^2} = 2\pi i \cdot 0 = 0$

Therefore, the value of the integral is $0$.

Would you like further details or have any questions?

Here are some additional questions you might have:

1. How do you determine the order of a pole in a complex function?
2. Can you explain how to compute residues for higher-order poles?
3. What are the applications of the residue theorem in physics and engineering?
4. How do you solve integrals with contours that enclose multiple poles?
5. Can you provide examples of integrals that do not result in zero?

Tip: Always verify the order of poles and apply the appropriate method for calculating residues, especially for higher-order poles.