91Ó°ÊÓ

The given function is \( f(z) = \frac{1}{(z-1)(z+2)^2} \). To find the poles, we set the denominator equal to zero: \((z-1)(z+2)^2 = 0\).This gives us the poles at \( z = 1 \) and \( z = -2 \). The pole \( z = 1 \) is simple, and \( z = -2 \) is a pole of order 2.

(a) For \(|z| = \frac{1}{2}\), both poles \(z = 1\) and \(z = -2\) lie outside the contour, so no poles are enclosed.(b) For \(|z| = \frac{3}{2}\), the pole \(z = 1\) is enclosed because its modulus is less than \( \frac{3}{2} \), but \(z = -2\) is still outside the contour.(c) For \(|z| = 3\), both poles are enclosed as their moduli (1 and 2) are less than 3.

The residue theorem states that \(\oint_{C} f(z) \, dz = 2\pi i \sum \text{Res}(f, z_i)\) where \(z_i\) are the poles inside \(C\).(a) Since \(|z| = \frac{1}{2}\) encloses no poles, the integral is zero.(b) For \(|z| = \frac{3}{2}\), only the residue at \(z=1\) is considered. For a simple pole at \(z = 1\), the residue is \(\text{Res}(f, 1) = \lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} \frac{1}{(z+2)^2} = \frac{1}{9}\).Thus, the integral is \(2 \pi i \times \frac{1}{9} = \frac{2 \pi i}{9}\).(c) For \(|z|=3\), both poles are enclosed. We already have \(\text{Res}(f, 1) = \frac{1}{9}\). For a pole of order 2 at \(z = -2\):\(\text{Res}(f, -2) = \lim_{z \to -2} \frac{d}{dz} \left( (z+2)^2 f(z) \right) = \lim_{z \to -2} \frac{d}{dz} \left( \frac{1}{z-1} \right) = -\frac{1}{9}\).Therefore, the integral is \(2 \pi i \times \left( \frac{1}{9} - \frac{1}{9} \right) = 0\).