Question

The value of \(\oint\) cos z dr where \(C\) is the circle \(|z|=1\), is .....

Short answer

The integral is 0.

Step-by-step solution

Understand the Problem

The problem asks for the value of a contour integral, specifically \( \oint_C \cos z \, dz \), where \( C \) is the circle \( |z| = 1 \). This is a closed contour integral of an analytic function \( \cos z \) over a path \( C \) that is a circle of radius 1 centered at the origin.

Review the Cauchy-Goursat Theorem

The Cauchy-Goursat Theorem states that if a function is analytic inside and on some closed contour \( C \), then the contour integral of the function over \( C \) is zero. \( \cos z \) is entire, meaning it is analytic everywhere on the complex plane.

Apply the Cauchy-Goursat Theorem

Since \( \cos z \) is analytic on and inside the circle \( |z| = 1 \), by the Cauchy-Goursat Theorem, the integral \( \oint_C \cos z \, dz \) is equal to zero.

Key concepts

Contour Integral

A contour integral in complex analysis is the integration of a complex function over a contour in the complex plane. Think of it as tracing the path of a curve and computing the cumulative effect of the function along this path. Contours are often closed curves, like circles, but can be more complex shapes. In our exercise, the contour is the circle defined by \( |z| = 1 \), meaning it has a radius of 1 and is centered at the origin of the complex plane.

This integral, \( \oint_C \cos z \, dz \), is evaluated around this circle. The concept of contour integration allows us to explore deeper properties of analytic functions in complex analysis. It's crucial to understand how these integrals behave, particularly in scenarios where analytic properties of the function apply to the contour.

Cauchy-Goursat Theorem

The Cauchy-Goursat Theorem plays a pivotal role in complex analysis, simplifying the computation of contour integrals. It states that if a function is analytic on and within a closed contour, the contour integral of that function over the contour is zero.

This theorem only applies when the function is free from singularities within the contour. In our problem, the function \( \cos z \) is analytic everywhere in the complex plane, including on the contour defined by \( |z| = 1 \). Thus, the contour integral of \( \cos z \) around this circle is zero.

The power of Cauchy-Goursat lies in its ability to collapse entire areas of potential computation into a straightforward answer, thus highlighting regions of analytic properties.

Analytic Function

Analytic functions are the backbone of complex analysis. These are functions which are locally expressible as a power series and differentiate smoothly everywhere within a domain. In simpler terms, analytic functions are those that behave nicely—they have derivatives at every point in their domain.

A function like \( \cos z \), involved in our problem, is called entire because it's analytic across the whole complex plane. Thus, it maintains its properties of continuity and differentiability everywhere, including on the circle \( |z| = 1 \) in our exercise.

Recognizing whether a function is analytic is crucial because it allows us to apply powerful theorems like the Cauchy-Goursat, simplifying otherwise complex integrals.

Complex Plane

The complex plane is the field on which the entire world of complex analysis functions. Visualize it as a two-dimensional plane composed of real and imaginary axes. The horizontal axis represents real numbers, while the vertical axis represents imaginary numbers. A complex number \( z = x + yi \) can be plotted on this plane with \( x \) being the real part and \( yi \) the imaginary part.

In our problem, the contour \( |z| = 1 \) represents a unit circle centered at the origin \( (0, 0) \) of this plane. It encompasses all points at an equal distance (a radius of one unit) from the origin.

Understanding the complex plane is essential when evaluating contour integrals as it provides a geometric interpretation of complex functions and the paths they traverse.