Let's solve the given contour integral step by step as requested: \[ \oint_{C} \frac{2z^3 + z^2 + 4}{z^4 + 4z^2} \, dz \] where \( C: |z-2| = 4 \) (clockwise direction). --- ### **Step 1: Simplify the integrand using partial fractions** #### **Explanation:** Factorize the denominator and express the integrand as a sum of simpler fractions. The denominator: \[ z^4 + 4z^2 = z^2(z^2 + 4) = z^2(z+2i)(z-2i) \] So, decompose: \[ \frac{2z^3 + z^2 + 4}{z^4 + 4z^2} = \frac{2z^3 + z^2 + 4}{z^2(z^2 + 4)} \] Let’s write: \[ \frac{2z^3 + z^2 + 4}{z^2(z^2 + 4)} = \frac{A}{z} + \frac{B}{z^2} + \frac{Cz + D}{z^2 + 4} \] Multiply both sides by \(z^2(z^2+4)\): \[ 2z^3 + z^2 + 4 = A z(z^2 + 4) + B(z^2 + 4) + (Cz + D)z^2 \] Expand: \[ = A(z^3 + 4z) + B(z^2 + 4) + Cz^3 + Dz^2 \] Group like terms: \[ 2z^3 + z^2 + 4 = (A+C)z^3 + (B+D)z^2 + 4A z + 4B \] Match coefficients: - \(z^3\): \(A + C = 2\) - \(z^2\): \(B + D = 1\) - \(z^1\): \(4A = 0\) ⇒ \(A = 0\) - Constant: \(4B = 4\) ⇒ \(B = 1\) Now, substitute \(A = 0\), \(B = 1\): - \(A + C = 2 \Rightarrow C = 2\) - \(B + D = 1 \Rightarrow 1 + D = 1 \Rightarrow D = 0\) So, the decomposition is: \[ \frac{2z^3 + z^2 + 4}{z^2(z^2 + 4)} = \frac{1}{z^2} + \frac{2z}{z^2 + 4} \] --- ### **Step 2: Evaluate the contour integral** #### **Explanation:** Integrate each term around the contour \(C\) (\(|z-2|=4\), clockwise). Use Cauchy's Integral Formula and the fact that the contour encloses \(z=0\) and \(z = \pm 2i\). - For \(\frac{1}{z^2}\): By Cauchy's Integral Formula (for derivative), \[ \oint_C \frac{1}{z^2}dz = 2\pi i \cdot f'(0) \] But \(f(z) = 1\), so \(f'(0) = 0\). Alternatively, recall: \[ \oint_C \frac{1}{(z-a)^n} dz = 0 \quad \text{for } n > 1 \] So, \(\oint_C \frac{1}{z^2} dz = 0\). - For \(\frac{2z}{z^2 + 4}\): Write as: \[ \frac{2z}{z^2 + 4} = \frac{2z}{(z-2i)(z+2i)} \] The poles \(z=2i\) and \(z=-2i\) are both inside the contour. Compute the residues at \(z=2i\) and \(z=-2i\): At \(z=2i\): \[ \text{Residue} = \lim_{z\to 2i} (z - 2i)\frac{2z}{(z-2i)(z+2i)} = \frac{2(2i)}{(2i+2i)} = \frac{4i}{4i} = 1 \] At \(z=-2i\): \[ \text{Residue} = \lim_{z\to -2i} (z + 2i)\frac{2z}{(z-2i)(z+2i)} = \frac{2(-2i)}{(-2i - 2i)} = \frac{-4i}{-4i} = 1 \] The sum of residues inside \(C\) is \(1 + 1 = 2\). Since the contour is clockwise, the integral gets a negative sign: \[ \oint_C \frac{2z}{z^2 + 4} dz = -2\pi i \cdot (\text{sum of residues}) = -2\pi i \times 2 = -4\pi i \] --- ### **Final Answer:** \[ \boxed{-4\pi i} \] **Explanation:** After decomposition, only the \(\frac{2z}{z^2 + 4}\) term contributes, and the sum of residues at poles \(z=2i\) and \(z=-2i\) is 2. The clockwise orientation gives a negative sign, resulting in \(-4\pi i\).