1. Problem statement: Evaluate the contour integrals given for various circles in the complex plane. 2. For ca) \(\oint_{|z|=1.5} \frac{dz}{z^2 - 2z} = \oint_{|z|=1.5} \frac{dz}{z(z-2)}\). Poles are at \(z=0\) and \(z=2\). Since \(|z|=1.5\), only \(z=0\) lies inside the contour. Residue at \(z=0\): \(\lim_{z\to0} z \cdot \frac{1}{z(z-2)} = \lim_{z\to0} \frac{1}{z-2} = -\frac{1}{2}\). By the residue theorem, integral = \(2\pi i \times (-\frac{1}{2}) = -\pi i\). 3. For cb) \(\oint_{|z|=1.5} \frac{dz}{2+z}\). Pole at \(z=-2\), which lies outside \(|z|=1.5\). No poles inside, integral = 0. 4. For c2) \(\oint \frac{z^2 - z + 1}{z-1} dz\). a) For \(|z|=1\): Pole at \(z=1\) lies on the contour, integral undefined or requires principal value. b) For \(|z|=0.5\): Pole at \(z=1\) lies outside contour, integral = 0. 5. For ③ around \(|z|=5\): a) \(\oint \frac{z^2 - 4z + 1}{z^2 - 2z} dz = \oint \frac{z^2 - 4z + 1}{z(z-2)} dz\). Poles at \(z=0\) and \(z=2\), both inside \(|z|=5\). Residue at \(z=0\): \(\lim_{z\to0} z \cdot \frac{z^2 - 4z + 1}{z(z-2)} = \lim_{z\to0} \frac{z^2 - 4z + 1}{z-2} = \frac{1}{-2} = -\frac{1}{2}\). Residue at \(z=2\): \(\lim_{z\to2} (z-2) \cdot \frac{z^2 - 4z + 1}{z(z-2)} = \lim_{z\to2} \frac{z^2 - 4z + 1}{z} = \frac{4 - 8 + 1}{2} = \frac{-3}{2} = -\frac{3}{2}\). Sum of residues = \(-\frac{1}{2} - \frac{3}{2} = -2\). Integral = \(2\pi i \times (-2) = -4\pi i\). b) \(\oint_{|z|=5} \frac{1}{z^2 + 1} dz = \oint_{|z|=5} \frac{1}{(z - i)(z + i)} dz\). Poles at \(z=i\) and \(z=-i\), both inside \(|z|=5\). Residue at \(z=i\): \(\lim_{z\to i} (z - i) \frac{1}{(z - i)(z + i)} = \frac{1}{2i}\). Residue at \(z=-i\): \(\lim_{z\to -i} (z + i) \frac{1}{(z - i)(z + i)} = \frac{1}{-2i} = -\frac{1}{2i}\). Sum of residues = \(\frac{1}{2i} - \frac{1}{2i} = 0\). Integral = 0. Final answers: - ca) \(-\pi i\) - cb) 0 - c2a) undefined (pole on contour) - c2b) 0 - ③a) \(-4\pi i\) - ③b) 0