Question

Evaluate the following integrals: a. \(\int_{C} \bar{z} d z\), where \(C\) is the parabola \(y=x^{2}\) from \(z=0\) to \(z=1+i\). b. \(\int_{C} f(z) d z\), where \(f(z)=2 z-\bar{z}\) and \(C\) is the path from \(z=0\) to \(z=2+i\) consisting of two line segments from \(z=0\) to \(z=2\) and then \(z=2\) to \(z=2+i\) c. \(\int_{C} \frac{1}{x^{2}+4} d z\) for \(C\) the positively oriented circle, \(|z|=2\). [Hint: Parametrize the circle as \(z=2 e^{i \theta}\), multiply numerator and denominator by \(e^{-i \theta}\), and put in trigonometric form.]

Short answer

The short answers for each integral are: a. \( \frac{3}{2} + i\), b. \(i\), c. \(π/2\)

Step-by-step solution

Problem a: Parametrization of the curve

For \(C\) being the parabola \(y= x^{2}\) from \(z=0\) to \(z=1+i\), one must parametrize the path along the real axis from \(0\) to \(1\) and along the imaginary axis from \(0\) to \(1\). This can be represented as \(z(t)=t\) for \(0 ≤ t ≤ 1\), and \(z(t) = 1+i(t-1)\) for \(1 ≤ t ≤ 2\), where \(t\) is a real parameter.

Problem a: Evaluation of the Integral

By substituting the parameterization and evaluating the definite integral, the solution can be found: \(\int_{C} \bar{z} d z\) can be divided into two parts. The first part \(\int_{0}^{1} \bar{t}dt = \frac{1}{2}\) and the second part \(\int_{1}^{2} \bar{1 + i(t-1)}dt = \frac{1}{2} + i\). So the integral is \(\frac{3}{2} + i\).

Problem b: Parameterization of the path

For \(C\) being the path from \(z=0\) to \(z=2+i\), one can parametrize the path along the real axis from \(0\) to \(2\) as \(z(t) = 2t\) for \(0 ≤ t ≤ 0.5\) and along the imaginary axis from \(2\) to \(2+i\) as \(z(t) = 2+i(4t-2)\) for \(0.5 ≤ t ≤ 1\)

Problem b: Evaluation of the Integral

Substitute the parameterization into the given integral and evaluate it. Firstly, \(\int_{0}^{0.5}2\cdot 2t - 2tdt = 0\). Secondly, \(\int_{0.5}^{1}2\cdot(2 + i(4t-2)) - 2-i(4t-2)dt = i\). So, the integral is \(i\).

Problem c: Parameterization of the circle

Here, the curve \(C\) is a positively oriented circle with \(|z| = 2\). Using the hint provided, it can be parameterized as \(z(t) = 2e^{it}\) where \(0 ≤ t ≤ 2π\) .

Problem c: Evaluation of the Integral

First multiply numerator and denominator by \(e^{-i t}\). This gives \(\frac{e^{-i t}}{4 + (2e^{-it})^{2}}\). After simplification, integral becomes \(\int_{0}^{2π}\frac{1}{4 + 4cos^{2}(t)}dt = π/2\). So the integral is \(π/2\).

Key concepts

Complex Integration

Complex integration is a vital aspect of complex analysis. It involves integrating complex-valued functions along a given path in the complex plane. This process is similar to real number integration but takes into account the intricate properties of complex numbers.

The integration is often performed by breaking a complicated path into simpler segments that are easier to manage. These segments can be straight lines or curves, and the integral across the entire path is obtained by summing up the integrals across each segment.

When evaluating integrals like those in the exercise, complex integration helps in calculating the function across specified paths with precision. Just like real line integrals, these integrals are path-dependent unless for specific conditions given by theorems like Cauchy’s Integral Theorem.

Parametrization

Parametrization is a fundamental technique used to express the path over which an integral is evaluated as a function of a single real variable, typically denoted as \(t\).

This powerful method simplifies complex integration by transforming the path into a more manageable set of functions. For example, curves and lines can be represented with equations involving \(t\), where \(z(t)\) denotes the path.

Consider the parabola where \(y=x^2\). It can be parametrized with \(z(t) = t + i t^2\) for an interval of \(t\). Parametrization breaks down such complex curves into simpler linear problems, easing the integration process.

Contour Integration

Contour integration is a technique employed specifically to evaluate integrals along paths, or contours, in the complex plane. This technique is immensely useful when dealing with complex functions that are otherwise difficult to integrate.

The contour integration method involves breaking down a complex path into simpler loops or curves. Then, the function is integrated along these paths, considering the orientation and shape of the contour. For example, four types of contours often considered are straight lines, circles, semicircles, and more intricate loops.

• In the case of a circle of radius 2, expressed as \(|z|=2\), parametrization like \(z(t) = 2e^{it}\) is used to facilitate integration around the circle.
• Performing contour integration effectively takes advantage of symmetry and complex variable properties to solve integrals efficiently.

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a cornerstone of complex analysis. It states that if a function is analytic (holomorphic) over a simply connected region, the integral of the function around any closed contour within that region is zero.

This theorem is extremely powerful as it allows the evaluation of integrals to be simplified significantly. The theorem implies that for any curve \(C\) that forms a closed path, the result of the integral \(\int_C f(z) \ dz\) is zero, provided \(f(z)\) is analytic in the region enclosed by \(C\).

Cauchy's theorem helps verify the independence of the path chosen for integration, facilitating complex integrations notably when verifying the solutions. This theory is foundational in proving various results in complex analysis and is instrumental when employing contour integration to solve intricate problems.