Question

Let \(C_{1}\) be a simple closed contour. Deform \(C_{1}\) into a new contour \(C_{2}\) in such a way that \(C_{1}\) does not encounter any singularity of an analytic function \(f\) in the process. Show that $$\oint_{C_{1}} f(z) d z=\oint_{C_{2}} f(z) d z$$ That is, the contour can always be deformed into simpler shapes (such as a circle) and the integral evaluated.

Short answer

The integral of function \(f\) over both the contours \(C_{1}\) and \(C_{2}\) is the same as per Cauchy's Theorem, which states that for a function analytic in a simply connected domain, the integral over any two contours in that domain will yield the same result, provided the contours can be deformed into each other without crossing any singularity. This confirms that contour integrals of analytic functions are path-independent and a contour can always be deformed into simpler shapes for computation without changing the result.

Step-by-step solution

Understanding the Concept of Deformation of Contours

The deformation of a contour involves transforming one contour to another without crossing any singularities of the function involved. It's vital to have a clear understanding of this concept as this forms the base for the problem.

Applying Cauchy's Theorem

An important theorem in analyzing contour integrals of analytic functions is Cauchy's Theorem. It states that if \(f\) is analytic over a simply connected domain \(D\) and \(C_{1}\) and \(C_{2}\) are contours lying in \(D\), then the integrals of \(f\) over \(C_{1}\) and \(C_{2}\) are equal. Since \(f\) is analytic and \(C_{1}\) is deformed to \(C_{2}\) without crossing any singularity, the conditions of the theorem are satisfied and we can assert that \(\oint_{C_{1}} f(z) d z=\oint_{C_{2}} f(z) d z\).

Conclusion

Thus, it's confirmed that the integral of an analytic function over two deformable contours that don't pass through any singularity of the function is the same. Deforming such a contour does not change the result of the integral. This conclusion aligns with the flexibility we have in choice of path while evaluating line integrals of analytic functions.

Key concepts

Complex Analysis

Complex analysis is a fascinating area of mathematics that studies functions involving complex numbers. These are the numbers of the form a + bi, where a and b are real numbers and i is the square root of -1. Complex analysis provides us with powerful tools to understand various properties of functions, including their integrals around certain paths or contours in the complex plane.

For instance, in our given problem, we focus on the integration along a path, or contour, in the complex plane. If you're picturing this, imagine a line that twists and turns in a space where every point represents a complex number. The integral we're addressing captures something about the function f(z) along this path. It's a bit like summing up how the function behaves over this journey, with complex analysis laying the rules for how this summing up works.

Cauchy's Theorem

Now, let's dive into one of the crown jewels of complex analysis - Cauchy's theorem. This revolutionary idea tells us that if we're working with an analytic function (one with a derivative at every point within a certain domain), and we're dealing with a simple, closed, deformable loop (contour) in this domain, the integral of the function around the loop is zero. This is a stunning realization because it means the value of the integral is not about the path; it's about the function and its analytic nature within the loop.

Applying this theorem to our exercise, we see why it doesn't matter if we reshape the contour, as long as we don't run into any singularities - the sharp points where the function goes 'wild' or is not defined. This concept ensures that \(\oint_{C_{1}} f(z) dz = \oint_{C_{2}} f(z) dz\) when properly transforming \( C_{1} \) into \( C_{2} \) without passing through any singularities.

Analytic Functions

Analytic functions, also known as holomorphic functions in complex analysis, are special because they are smooth and well-behaved. They have derivatives everywhere in their domain, which makes them predictable and easily manageable mathematically. When we talk about a function being analytic, we basically mean that it can be expressed as a power series - an infinite sum akin to a polynomial but with potentially infinitely many terms.

For our exercise, the fact that \( f \) is an analytic function is crucial because it ensures that applying Cauchy's theorem is legitimate. Analyticity guarantees no abrupt changes as we move the contour \( C_{1} \) around, as long as we avoid the trouble spots - the singularities.

Singularities in Complex Functions

The final core concept you need to understand for the exercise is singularities in complex functions. Singularities are points where a function doesn't behave nicely - it might go to infinity, or it's just not defined there. They're like obstacles in your path in the complex plane.

When solving the exercise, avoiding these singularities while deforming the contour \( C_{1} \) to \( C_{2} \) is like making sure your path doesn't cross any 'no-go' zones. If you accidentally stepped into these zones while reshaping the contour, you couldn't apply Cauchy's theorem, and the resulting integrals might be different. The singularity is essentially where the function 'breaks' the rules that apply to its nicer, analytic neighbors.