Let \(z\) be a complex variable. If \(f(z) = \frac{\sin(\pi z){z^2 (z - 2)}\) and \(C\) is the circle in the complex plane with \(|z| = 3\), then:}
\[
\oint_C f(z) \, dz
\]
is \(\_\_\_\_\).
Show Hint
For contour integrals, use the residue theorem by identifying the poles, determining their residues, and summing the contributions inside the contour.
Step 1: Locate the poles of \(f(z)\).
The function \(f(z)\) has two poles inside the contour \(|z| = 3\):
At \(z = 0\) with order 2.
At \(z = 2\) with order 1.
Step 2: Use the residue theorem.
According to the residue theorem, the contour integral is:
\[
\oint_C f(z) \, dz = 2\pi j \left({Residue at } z = 0 + {Residue at } z = 2\right).
\]
Step 3: Compute the residues.
For \(z = 0\):
\[
{Residue at } z = 0 = \lim_{z \to 0} \frac{d}{dz} \left(z^2 f(z)\right) = -\pi^2.
\]
For \(z = 2\):
\[
{Residue at } z = 2 = \frac{\sin(\pi \cdot 2)}{2^2} = \frac{\sin(2\pi)}{4} = 0.
\]
Step 4: Evaluate the integral.
Substituting the residues into the formula:
\[
\oint_C f(z) \, dz = j \cdot (-\pi^2) = -\pi^2 j.
\]
Final Answer:
\[\boxed{{(4) } -\pi^2 j}\]