Question

Use Cauchy's theorem or integral formula to evaluate the integrals. $${\oint }_C\frac{\sin zdz}{2z-\pi }\text{ where }C\text{ is the circle }\begin{array}{l}\text{ (a) }|z|=1\\ \text{ (b) }|z|=2\end{array}$$

Short answer

For part (a), the integral is 0. For part (b), the integral is $2\pi i$.

Step-by-step solution

Understanding the Problem

Given the integral ${\oint }_C\frac{\sin zdz}{2z-\pi }$, Cauchy's theorem or integral formula can be used to evaluate it over different contours C.

Identify the nature of the integrand's singularity

The integrand $\frac{\sin z}{2z-\pi }$ has a singularity where the denominator equals zero. Solve $2z-\pi =0$ to find the singularity at $z=\frac{\pi }{2}$.

Evaluate for $|z|=1$

For part (a), the singularity $z=\frac{\pi }{2}\approx 1.57$ lies outside the contour $|z|=1$. Therefore, by Cauchy's theorem, since there are no singularities inside the contour, the integral is zero.

Evaluate for $|z|=2$

For part (b), the singularity $z=\frac{\pi }{2}$ lies within the contour $|z|=2$. According to the Cauchy integral formula, the integral ${\oint }_C\frac{f(z)}{z-z_0}dz=2\pi if(z_0)$ where $f(z)=\sin z$ and $z_0=\frac{\pi }{2}$. Therefore, the integral is: $${\oint }_{|z|=2}\frac{\sin zdz}{2z-\pi }=2\pi i\cdot \sin \left(\frac{\pi }{2}\right)\text{, where }\sin \left(\frac{\pi }{2}\right)=1$$ This yields: $2\pi i\times 1=2\pi i$.

Key concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. These functions often exhibit interesting properties not seen in real-valued functions. For example, complex functions can have singularities, points where they become undefined or infinite.

Key concepts in complex analysis include:

• Complex Differentiation: Functions that are differentiable in the complex sense. These functions are called holomorphic or analytic.
• Contour Integration: Integrating complex functions along a path in the complex plane.
• Singularities: Points where a function does not behave well, such as poles or essential singularities.
• Residue Theorem and Cauchy's Theorems: Tools to evaluate complex integrals based on the behavior around singularities.

This branch provides powerful tools to solve many integrals, differential equations, and other problems that are difficult to tackle using real analysis alone.

Contour Integration

Contour integration involves evaluating complex integrals over a specified path or contour in the complex plane. A contour is usually a closed curve, and the integral is taken around this curve.

Steps for contour integration:

• Select Contour: Choose the path over which to integrate. This is often a circle or rectangle for simplicity.
• Parameterize the Contour: Express the contour in terms of a parameter, usually an angle for circles.
• Identify Singularities: Determine where the function being integrated has singularities relative to the contour.
• Apply Cauchy's Theorem: If no singularities are inside the contour, the integral is zero.
• Use Residue Theorem or Integral Formula: If singularities are inside the contour, use these theorems to evaluate the integral.

In simple terms, contour integration leverages the geometry of the complex plane and the properties of analytic functions to simplify and solve integrals.

Singularity in Complex Functions

A singularity in a complex function is a point where the function ceases to be well-behaved in some particular way, such as becoming infinite or undefined. Singularities are crucial in complex analysis because they often dictate the behavior of integrals and other properties of the function.

Types of Singularities:

• Pole: A point where the function approaches infinity. The order of the pole tells us how the function behaves near this point.
• Removable Singularity: A point where the function could be redefined to make it analytic. For instance, a hole in the function.
• Essential Singularity: A point where the function exhibits erratic behavior near it, with neither a pole nor a removable singularity.

In our example integral, the function $\frac{\sin z}{2z-\pi }$ has a singularity at $z=\frac{\pi }{2}$ which is a simple pole. This singularity influences whether the contour integral evaluates to zero or not, based on its location in relation to the contour.

Cauchy Integral Formula

The Cauchy Integral Formula provides a powerful way to evaluate integrals of analytic functions. It says that for a function $f(z)$ analytic inside and on a simple closed contour $C$ containing a point $z_0$, the formula is:

$$f(z_0)={\oint }_C\frac{f(z)}{z-z_0}dz=2\pi if(z_0)$$

This remarkable formula not only allows us to find the value of the function at a point but also evaluate integrals of functions that can be written in that form.

Key Aspects:

• Analytic Function: The function $f(z)$ must be analytic within and on the contour $C$.
• Singularity Inside Contour: The point $z_0$ must be inside the contour $C$ for the formula to apply.
• Integral Format: The integral format ${\oint }_C\frac{f(z)}{z-z_0}dz$ is crucial as it utilizes the singularities directly in a simple pole form.

In the exercise given, since $\text{Missing argument for frac}$ has a simple pole at $z=\frac{\pi }{2}$, we use Cauchy's Integral Formula to evaluate the integral in part (b), which lies within the chosen contour.