Let's break the problem into steps and answer each as requested: --- ### (a) **Sketch the Contour \( C(R) \) and Indicate Singularities** #### **Step 1: Identify the contour and singularities** - The contour \( C(R) \) consists of two parts: 1. The line segment from \( -R \) to \( R \) on the real axis. 2. The semicircular arc \( C_R \) of radius \( R \) in the upper half-plane, parameterized by \( Re^{it} \), \( t \in [0, \pi] \). - The function \( f(x) = \frac{xe^{i4x}}{x^2 - 2x + 2} \) has singularities where the denominator is zero: \[ x^2 - 2x + 2 = 0 \implies x = 1 \pm i \] So, there are simple poles at \( z = 1+i \) and \( z = 1-i \). #### **Explanation:** - The contour \( C(R) \) is a closed semicircular path in the upper half-plane, oriented counterclockwise. - The singularity \( z = 1+i \) lies inside the contour if \( R > \sqrt{2} \); \( z = 1-i \) is below the real axis and outside the contour. --- ### (b) **Value of \( \int_{C(R)} f(z)\,dz \) for \( R < \sqrt{2} \)** #### **Step 2: Analyze the contour for \( R < \sqrt{2} \)** - For \( R < \sqrt{2} \), the contour does **not** enclose any singularity (since both \( 1+i \) and \( 1-i \) have modulus \( \sqrt{2} \)). - By Cauchy's theorem, the integral around a closed contour that does **not** enclose any singularities is **zero**. #### **Explanation:** - Since the contour does not enclose any poles, the integral is zero by Cauchy's theorem. --- ### **Final Answers for Steps (a) and (b):** **(a) Answer:** - The contour \( C(R) \) is a closed path consisting of a line segment from \( -R \) to \( R \) (real axis), and a semicircular arc of radius \( R \) in the upper half-plane (from \( R \) back to \( -R \)), oriented counterclockwise. The only singularity inside the contour for \( R > \sqrt{2} \) is \( z = 1 + i \). **(b) Answer:** - For \( R < \sqrt{2} \), \( \int_{C(R)} f(z)\,dz = 0 \), because the contour does not enclose any singularities. --- Let me know if you want to proceed to steps (c), (d), and (e)!

Certainly! Continuing with the remaining parts: --- ### **(c) Show that \(\displaystyle \oint_{C(R)} f(z) \, dz = 2\pi i \times \text{Res}_{z=1+i} f(z)\) for \( R > \sqrt{2} \)** #### **Step 3: Apply the Residue Theorem** - Since the contour \( C(R) \) encloses the pole at \( z=1+i \) when \( R > \sqrt{2} \), by the Residue Theorem: \[ \oint_{C(R)} f(z) \, dz = 2\pi i \times \operatorname{Res}_{z=1+i} f(z) \] - The residue at \( z=1+i \): --- ### **(d) Compute \( \operatorname{Res}_{z=1+i} f(z) \)** - The function: \[ f(z) = \frac{z e^{i4z}}{z^2 - 2z + 2} \] - Denominator factors as: \[ z^2 - 2z + 2 = (z - (1+i))(z - (1-i)) \] - The pole at \( z=1+i \) is simple. Residue at \( z=1+i \): \[ \operatorname{Res}_{z=1+i} f(z) = \lim_{z \to 1+i} (z - (1+i)) \frac{z e^{i4z}}{(z - (1+i))(z - (1-i))} = \lim_{z \to 1+i} \frac{z e^{i4z}}{z - (1-i)} \] - Substitute \( z = 1+i \): \[ \operatorname{Res}_{z=1+i} f(z) = \frac{(1+i) e^{i4(1+i)}}{(1+i) - (1-i)} = \frac{(1+i) e^{i4(1+i)}}{(1+i) - (1-i)} \] - Simplify denominator: \[ (1+i) - (1-i) = i + i = 2i \] - Simplify numerator: \[ e^{i4(1+i)} = e^{i4 + i4i} = e^{i4} \times e^{i4i} \] But \( i4i = i \times 4i = 4i^2 = 4 \times (-1) = -4 \), so: \[ e^{i4(1+i)} = e^{i4} \times e^{-4} \] - Final residue: \[ \operatorname{Res}_{z=1+i} f(z) = \frac{(1+i) e^{i4} e^{-4}}{2i} \] --- ### **(e) Final expression:** \[ \boxed{ \oint_{C(R)} f(z) \, dz = 2\pi i \times \frac{(1+i) e^{i4} e^{-4}}{2i} = \pi \times (1+i) e^{i4} e^{-4} } \] which simplifies to: \[ \boxed{ \pi (1 + i) e^{i4 - 4} } \] --- ### **Summary:** **(c)** For \( R > \sqrt{2} \), the integral over \( C(R) \) equals \( 2\pi i \) times the residue at \( z=1+i \). **(d)** The residue at \( z=1+i \): \[ \operatorname{Res}_{z=1+i} f(z) = \frac{(1+i) e^{i4} e^{-4}}{2i} \] **(e)** Final value of the integral: \[ \boxed{ \pi (1 + i) e^{i4 - 4} } \] --- Let me know if you'd like further clarifications!