Problem 7 - Advanced Engineering Mathematics

Question

Prove that \(\oint_{c} f(z) d z=0\), where \(f\) is the given function and \(C\) is the unit circle \(|z|=1\). \(f(z)=\tan z\)

Step-by-Step Solution

Answer

The integral is zero since \(\tan z\) is analytic on the unit circle.

Step 1: Cauchy's Theorem for Contours

To solve this exercise, we begin by invoking Cauchy's Theorem, which states that if a function is analytic within and on a closed contour, the integral over this contour is zero. Thus, we need to check if \(f(z) = \tan z\) is analytic on the unit circle \(|z| = 1\).

Step 2: Identifying Singularities

The function \(f(z) = \tan z\) has singularities (poles) where \(\cos z = 0\). Since \(\cos z = 0\) at \(z = \frac{\pi}{2} + n\pi\), where \(n\) is an integer, these points are singularities of the function. We need the contour \(|z| = 1\) to avoid passing through any of these poles.

Step 3: Determining Singularities on the Contour

We check for which \(n\), the singularities \(z = \frac{\pi}{2} + n\pi\) lies on the unit circle \(|z| = 1\). For the unit circle, \(|z| = 1\) is equivalent to finding real values of \(z\) such that \(\mathrm{Re}(e^{iz}) = 1\) and \(\mathrm{Im}(e^{iz}) = 0\). None of \(\frac{\pi}{2} + n\pi\) lie on the unit circle for integer \(n\) within the domain \([0, 2\pi)\).

Step 4: Application of Cauchy's Theorem

Since there are no singular points of \(f(z) = \tan z\) on or within the unit circle \(|z| = 1\), \(f(z)\) is analytic on the unit circle. By Cauchy's theorem, we conclude that the integral of a function analytic on and within a closed contour is equal to zero.

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that deals with functions of complex numbers. In our exercise, we're working with functions like \(f(z) = \tan z\), where \(z\) is a complex number of the form \(a + bi\) (\(a\) and \(b\) are real numbers). The beauty of complex analysis lies in its ability to capture the behavior of these functions within the complex plane.

• Complex analysis examines both the algebraic and geometric aspects of complex numbers.
• One of its key tools is the complex integral, which involves integrating a complex function along a path or contour in the complex plane.

In the exercise, we're asked to show that the contour integral of a specific complex function is zero, which is a typical problem in this field. This task involves understanding the interplay between the function and the geometry of its domain.

Analytic Function

An analytic function, also known as a holomorphic function, is a function that is locally represented by a convergent power series. In simple terms, these functions are very smooth and have derivatives of all orders.

• Analytic functions are defined on and within a certain region, meaning they can't have any gaps or jumps within this area.
• They possess a derivative at every point in their domain.

Analyzing whether \(f(z) = \tan z\) is analytic is crucial to applying Cauchy's theorem. Specifically, a function must have no singularities within the contour for its integral to equal zero, implying the function is analytic in that region.

Unit Circle

The unit circle is a circle with a radius of one, centered at the origin of the complex plane. When working with a contour integral, as in this exercise, the contour \(|z| = 1\) refers to the path traced by the unit circle.

• Points on the unit circle satisfy the equation \(|z| = 1\).
• It is often used in complex analysis due to its simplicity and symmetry.

In our problem, we're interested in finding if any singularities of \(f(z) = \tan z\) occur on this circle. The absence of these singular points within and on the unit circle indicates the function might be analytic, thereby allowing the use of Cauchy's theorem.

Singularity

A singularity in complex analysis refers to a point where a function does not behave as expected, such as having an undefined derivative or infinite value. Identifying singularities is essential in determining where a function ceases to be analytic.

• For \(f(z) = \tan z\), singularities occur where \(\cos z = 0\).
• These are important because they determine the regions where the function cannot be considered analytic.

In our solution, we identified that \(z = \frac{\pi}{2} + n\pi\) are singular points, but none of these lie on the unit circle \(|z| = 1\). Thus, \(f(z)\) is analytic along this path, allowing us to conclude the contour integral is zero.