Question

Let \(C\) be the boundary of the square with vertices at the points \(z=0\), \(z=1, z=1+i\), and \(z=i\) with counterclockwise direction. Evaluate $$\oint_{C}(5 z+2) d z \text { and } \oint_{C} e^{\pi z^{*}} d z$$

Short answer

The result of the first integral is \(0\). The result of the second integral can be found by doing the calculations described in the step 'Sum the integrals'.

Step-by-step solution

Evaluate the first integral

Due to Cauchy's integral theorem, the integral of a holomorphic function along a simple closed curve equals zero. Since \(5z + 2\) is a holomorphic function, it holds that \(\oint_{C}(5 z+2) d z = 0\).

Parametrize the path

To evaluate the second line integral, parametrize the boundary \(C\) of the square. The four segments from \(z = 0\) to \(z=1\), from \(z=1\) to \(z = 1 + i\), from \(z = 1 + i\) to \(z = i\), and from \(z = i\) to \(z = 0\) can be parametrized as \(z_1(t) = t\), \(z_2(t) = 1 + it\), \(z_3(t) = 1 + (1 - t)i\), and \(z_4(t) = (1 - t)i\), respectively, with \(t \in [0, 1]\).

Calculate each integral

The second integral is subsequently found by calculating the line integral along each segment: \[\oint_{C} e^{\pi z^{*}} d z = \int_{0}^{1} e^{\pi (0)} dz_1 + \int_{0}^{1} e^{\pi (1 - it)} dz_2 + \int_{0}^{1} e^{\pi (1+ it)} dz_3 + \int_{0}^{1} e^{\pi (it)} dz_4\].

Sum the integrals

In order to find the value of the total integral, compute the four integrals and sum them up.

Key concepts

Cauchy's Integral Theorem

Cauchy's integral theorem is a fundamental result in complex analysis that provides a powerful tool for evaluating certain integrals. It states that if a function is holomorphic (complex differentiable) on and inside a closed curve, then the integral of this function over the curve is zero. This theorem requires that the function does not have any singularities (points where it is not defined) inside the curve. Given its implications, this theorem simplifies many calculations related to complex integrals.

In the provided exercise, the integral of the function \(5z + 2\) over the closed curve \(C\) is zero because \(5z + 2\) is indeed a holomorphic function on the complex plane. Knowing this, we can quickly conclude that the integral along the curve results in zero without performing further computations. Understanding this principle can greatly simplify solving many complex integral problems.

Holomorphic Functions

A function is said to be holomorphic on a region if it is complex differentiable at every point within that area. This requirement is stronger than the real differentiability because it demands the function to be differentiable with respect to complex variables, meaning both real and imaginary parts satisfy the Cauchy-Riemann equations.

In practice, holomorphic functions can be thought of as complex analogs to smooth, continuously differentiable functions. They exhibit nice properties, such as continuous derivatives, and play a pivotal role in complex analysis. Functions like \(5z + 2\) fall under this classification, which we identify in the exercise. It assures us that certain theorems, like Cauchy's, hold true, allowing elegant solutions to seemingly complicated problems.

Line Integrals

Line integrals in complex analysis involve integrating a complex function over a path or curve in the complex plane. Unlike real integrals, these integrals take into account both the magnitude and the direction of the path. They play an essential role in applications such as fluid dynamics, electromagnetism, and in solving partial differential equations.

The calculation generally follows this process: break down the curve into segments, integrate over each segment, and then sum the results. To solve for the integral \(\oint_{C} e^{\pi z^{*}} dz\), the curve is segmented, making the process manageable by solving smaller integrals over each path segment. By understanding line integrals, one can grasp more complex phenomena described in advanced mathematics.

Parametrization of Curves

Parametrization is the process of defining a path using a single parameter, typically denoted as \(t\). It allows for the straightforward evaluation of line integrals by providing a clear definition of how a complex path is traversed.

In the original exercise, the boundary curve \(C\) of the square is parametrized using straightforward linear functions describing each side of the square. For instance, one segment from \(z = 0\) to \(z = 1\) can be parametrized as \(z_1(t) = t\), and similarly for the other sides of the square. This approach converts what could be a complex dependency into simpler, isolated integrals over \([0, 1]\). Understanding how to parametrize allows for practical solutions to integral evaluations and is a critical skill when dealing with curves in any dimension.