Problem 7 - Complex Analysis for Mathematics and Engineering

Question

Recall \(C_{r}^{+}(a)\) is the circle of radius \(r\) centered at \(a\), oriented eounterclockwise. (a) Evaluate \(\int_{C_{i}^{+}(0)} z d z\). (b) Evaluate \(\int_{\sigma_{4}^{+}(0)} \bar{y} d z\). (c) Evaluate \(\int_{C_{2}^{-}(0)} \frac{1}{x} d z\). (The minus sign means elockwise orientation.) (d) Evaluate \(\int_{C_{2}^{-}(0)} \frac{1}{2} d z\). (e) Evaluate \(\int_{C}(z+1) d z\), where \(C\) is \(C_{1}^{+}(0)\) in the first quadrant. (f) Evaluate \(\int_{c}\left(x^{2}-i y^{2}\right) d z\), where \(C\) is the upper half of \(C_{1}^{+}(0)\). (g) Evaluate \(\int_{C}|z-1|^{2} d z\), where \(C\) is the upper half of \(C_{1}^{+}(0)\).

Step-by-Step Solution

Answer

Solutions: (a) 0, (b) 32, (c) -2πi, (d) -2π, (e) πi/2, (f) 0, (g) 0.

Step 1: Evaluate (a) Integral

We need to evaluate the integral \( \int_{C_{i}^{+}(0)} z \, dz \). The integrand \( z \) (which is the complex variable itself) and the contour \( C_{r}^{+}(a) \) is a path that loops around the origin once counterclockwise. According to Cauchy's Integral Theorem, if \( z \, dz \) parameterizes a closed contour across a path with no singularities inside, it simplifies to zero:\[ \int_{C_{r}^{+}(a)} z \, dz = 0 \] Thus, the solution to (a) is 0.

Step 2: Evaluate (b) Integral

For \( \int_{\sigma_{4}^{+}(0)} \bar{y} \, dz \), since the integral involves \( \bar{y} \), replace \( \bar{y} \) with \( y \). \( \sigma_{4}^{+}(0) \) represents a semicircular path above the real axis. Parameterize it using \( z = 4e^{i\theta} \), with \( 0 \leq \theta \leq \pi \). Decompose \( z = x + iy = 4 \cos \theta + i 4 \sin \theta \). Then, \( \bar{y} = -iy = -i(4 \sin \theta) \). The differential \( dz = 4ie^{i\theta} \, d\theta \).Thus, the integral becomes:\[ \int_{0}^{\pi} -4i^2 \sin \theta \cdot 4ie^{i\theta} \, d\theta = \int_{0}^{\pi} 16 \sin \theta \, d\theta \]Evaluating \( \int_{0}^{\pi} 16 \sin \theta \, d\theta = -16(-\cos \theta)|_{0}^{\pi} = -16(-(-1) - 1) = 32 \).So the solution to (b) is 32.

Step 3: Evaluate (c) Integral

For \( \int_{C_{2}^{-}(0)} \frac{1}{x} \, dz \), note that "clockwise" makes the integral negative. Rewrite \( \frac{1}{x} \) as \( \frac{1}{z} \) and apply Cauchy's Integral Theorem:\[ \int_{C_{2}^{-}(0)} \frac{1}{z} \, dz = -2\pi i \]Hence the answer for (c) is \(-2\pi i\).

Step 4: Evaluate (d) Integral

Since \( \frac{1}{2} \) is constant, \( \int_{C_{2}^{-}(0)} \frac{1}{2} \, dz \) becomes \( \frac{1}{2} \times \text{length of } C_{2}^{-}(0) \). Circle \( C_{2}^{-}(0) \) has radius 2:Circumference = \( 2\pi \times 2 = 4\pi \)Thus, \(-4\pi \times \frac{1}{2} = -2\pi \).Therefore, (d) evaluates to \(-2\pi \).

Step 5: Evaluate (e) Integral

For \( \int_{C}(z+1) \, dz \), where \( C \) is \( C_{1}^{+}(0) \), parameterize quarter-circle contour: \( z = e^{i\theta} \); from \( 0 \leq \theta \leq \frac{\pi}{2} \). Convert \( z+1 = e^{i\theta} + 1 \).Calculate using integral symmetry:\[ \int(z+1) \, dz = \int e^{i\theta} e^{i\theta}i \, d\theta + \int i \, d\theta = \left[ \frac{i e^{2i\theta}}{2} + i\theta \right]_{0}^{\pi/2} \]Integrating yields: \( \left[ \frac{i}{2}((-1) - (1)) + i\left(\frac{\pi}{2}\right) \right] = \frac{\pi i}{2} \).Thus, the solution for (e) is \( \frac{\pi i}{2} \).

Step 6: Evaluate (f) Integral

For \( \int_{c}(x^{2} - i y^{2}) \, dz \) on half \( C_{1}^{+}(0) \), parameterize semicircle in top half: \( x^{2} - i y^{2} \rightarrow (\cos^{2}\theta - i\sin^{2}\theta) \, i e^{i\theta} \). Find derivative \( dz = ie^{i\theta} \, d\theta \).Integrate:\[ \int_{0}^{\pi} (\cos^2 \theta - i \sin^2 \theta) i e^{i\theta} \frac{\pi}{2} \, d\theta \]Answer gives \( \left[ \sin \theta \right]_{0}^{\pi} = 0 \).Thus, (f) this evaluates to 0.

Step 7: Evaluate (g) Integral

Evaluate \( \int_{C}|z-1|^{2} \, dz \) over top half \( C_{1}^{+}(0) \). Simplify using parameterization \( |z-1| = |e^{i\theta} - 1|^2 = 2 - 2\cos \theta \). Integral becomes:\[ \int_{0}^{\pi} (2 - 2\cos \theta) \, ie^{i\theta} \, d\theta \]Integrate circular path:\( 2i \int e^{i\theta} \, d\theta = \int (2ie^{i\theta} + ie^{i\theta} ) \, d\theta \) and zero out on complete loop via symmetry cancelling cosine terms.Thus, the solution evaluates to 0 for (g).

Key Concepts

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a central result in complex analysis, which states that if a function is analytic and has no singularities within a closed contour in a complex plane, then the contour integral of this function around that contour is zero. This theorem has powerful implications:

• An analytic function is also known as a holomorphic function, meaning it has complex derivatives at every point in its domain.
• Singularities are points where the function is not analytic, such as points of division by zero.
• A closed contour simply means that the path starts and ends at the same point without intersecting itself, like a loop.

This theorem helps us evaluate integrals by confirming if the function is analytic within and on the contour, leading often to the result being zero, as demonstrated in parts (a) and (f) of the original exercise.

Contour Integrals

Contour integrals are integrals taken over a path or curve in the complex plane.

• They differ from typical real integrals by integrating a complex function along a path defined by the contour.
• The direction of the contour (clockwise or counterclockwise) influences the sign of the integral. For instance, a clockwise path often leads to a negative value, as seen in part (c) of the exercise.
• Parameterization is key in evaluating these integrals effectively, where the variable change helps calculate the integral over the specified path.

These integrals are crucial in evaluating complex functions over closed paths, especially when invoking theorems like Cauchy's Integral Theorem to simplify the evaluation.

Complex Integration

Complex integration involves integrating complex-valued functions, often done over contours or paths in the complex plane.

• The integrand can involve various expressions including polynomials and fractions, often involving complex conjugates such as in parts (b) and (d) of the exercise.
• Integration is performed by parameterizing the contour, transforming the complex function in terms of a single real variable, such as using polar coordinates for a circular path.
• These techniques facilitate solving complex integrals, taking advantage of symmetry and analytic properties of the functions involved.

Understanding complex integration is pivotal in complex analysis, enabling one to handle a wide range of complex functions within mathematical physics and engineering.