Let's solve each part step by step with explanations: --- ## **Ba) Half-Disk Integral of \( f(z) = \sqrt{r} e^{i\theta/2} \)** ### **Step 1: Understand the Contour \( C \) and the Function** - The contour \( C \) is the boundary of the half-disk: \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq \pi \). - \( f(z) \) is defined as \( f(0) = 0 \) and elsewhere as \( f(z) = \sqrt{r} e^{i\theta/2} \), which is a branch of \( z^{1/2} \). **Explanation:** The function is continuous on the half-disk, and the contour runs along the semicircle from \( \theta = 0 \) to \( \theta = \pi \) and back along the diameter. --- ### **Step 2: Parameterize the Contour** - Along the semicircular arc (\( r = 1 \), \( 0 \leq \theta \leq \pi \)): - \( z = e^{i\theta} \), \( dz = i e^{i\theta} d\theta \) - \( f(z) = e^{i\theta/2} \) - Along the diameter (\( \theta = 0 \)): - \( z = x \), \( -1 \leq x \leq 1 \) - \( f(z) = \sqrt{x} \) for \( x \geq 0 \), not defined for \( x < 0 \), but since \( f(0) = 0 \), the integrand is 0 along the diameter. **Explanation:** The integral along the diameter is zero because \( f(z) = 0 \) there. So, only the semicircular arc contributes. --- ### **Step 3: Compute the Integral on the Semicircular Arc** \[ \int_C f(z) dz = \int_{0}^{\pi} e^{i\theta/2} \cdot i e^{i\theta} d\theta = i \int_{0}^{\pi} e^{i\theta(1 + 1/2)} d\theta = i \int_{0}^{\pi} e^{i\frac{3\theta}{2}} d\theta \] Integrate: \[ i \int_{0}^{\pi} e^{i\frac{3\theta}{2}} d\theta = i \left[ \frac{2}{3i} e^{i\frac{3\theta}{2}} \right]_{0}^{\pi} = \frac{2}{3} \left(e^{i\frac{3\pi}{2}} - 1\right) \] \( e^{i\frac{3\pi}{2}} = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right) = 0 - i = -i \) So: \[ \frac{2}{3}(-i - 1) = -\frac{2}{3}(1 + i) \] **Explanation:** By parameterizing and integrating, we find the value of the contour integral over the semicircular arc. --- ### **Final Answer for Ba:** \[ \boxed{ \int_C f(z) dz = -\frac{2}{3}(1 + i) } \] --- ## **Sa) Two Contour Integrals over a Square** ### **Step 1: Analyze the Poles Inside the Square** - The square: \( x, y = \pm 2 \) (so \( |z| \leq 2\sqrt{2} \)) - For \( \int_C \frac{\cos z}{z(z^2 + 8)} dz \): - Poles at \( z = 0, z = \pm 2\sqrt{2}i \) - All are inside the square. - For \( \int_C \frac{\cosh z}{z^4} dz \): - Pole of order 4 at \( z = 0 \) **Explanation:** Identifying the singularities inside the contour is crucial for using the residue theorem. --- ### **Step 2: Compute the Residues** #### **First Integral:** \[ f(z) = \frac{\cos z}{z(z^2 + 8)} \] - At \( z=0 \): Residue is \( \lim_{z \to 0} \frac{\cos z}{z^2 + 8} = \frac{1}{8} \) - At \( z = 2\sqrt{2}i \): Simple pole, so residue is \( \frac{\cos(2\sqrt{2}i)}{2\sqrt{2}i [2\sqrt{2}i + 0]} \). But \( \cosh(2\sqrt{2}) \) in numerator. - Similarly for \( z = -2\sqrt{2}i \). #### **Second Integral:** \[ g(z) = \frac{\cosh z}{z^4} \] - Only pole at \( z=0 \), order 4. - Residue is \( \frac{1}{3!} g'''(0) \) - \( \cosh z = 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \ldots \) - \( g(z) = \frac{1}{z^4} + \frac{1}{2! z^2} + \frac{1}{4!} + \ldots \) - Coefficient of \( 1/z \) is 0. **Explanation:** For higher order poles, Laurent expansion or derivative formulas are used. --- ### **Step 3: Apply the Residue Theorem** - Each integral equals \( 2\pi i \) times the sum of the residues inside \( C \). #### **First Integral:** \[ \text{Residue at } z = 0: \frac{1}{8} \] \[ \text{Residue at } z = 2\sqrt{2}i: \frac{\cos(2\sqrt{2}i)}{2\sqrt{2}i [2\sqrt{2}i - 0]} = \frac{\cosh(2\sqrt{2})}{-8} \] \[ \text{Residue at } z = -2\sqrt{2}i: \frac{\cos(-2\sqrt{2}i)}{-2\sqrt{2}i [-2\sqrt{2}i - 0]} = \frac{\cosh(2\sqrt{2})}{-8} \] Sum: \( \frac{1}{8} - \frac{\cosh(2\sqrt{2})}{4} \) So, \[ \int_C \frac{\cos z}{z(z^2+8)}dz = 2\pi i \left(\frac{1}{8} - \frac{\cosh(2\sqrt{2})}{4}\right) \] #### **Second Integral:** - Coefficient of \( 1/z \) in Laurent expansion is 0. So, \[ \int_C \frac{\cosh z}{z^4}dz = 0 \] **Explanation:** Residue theorem gives the answer as \( 2\pi i \) times the sum of residues. --- ### **Final Answers for Sa:** \[ \boxed{ \int_C \frac{\cos z}{z(z^2 + 8)}dz = 2\pi i \left(\frac{1}{8} - \frac{\cosh(2\sqrt{2})}{4}\right) } \] \[ \boxed{ \int_C \frac{\cosh z}{z^4}dz = 0 } \] --- **Summary Table:** | Part | Final Answer | |------|--------------| | Ba | \( -\frac{2}{3}(1 + i) \) | | Sa (1) | \( 2\pi i \left(\frac{1}{8} - \frac{\cosh(2\sqrt{2})}{4}\right) \) | | Sa (2) | 0 |