Problem 1 - 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 \(|z|=1\). \(f(z)=z^{3}-1+3 i\)

Step-by-Step Solution

Answer

The integral \(\oint_{C} f(z) \, dz = 0\) because \(f(z)\) is analytic on and inside the unit circle.

Step 1: Identify the Type of Integral

We are given a contour integral of the form \(\oint_{C} f(z) \, dz\) where \(C\) is the unit circle \(|z|=1\). Since the contour is closed and \(f(z)\) is a polynomial, we need to verify if the conditions for Cauchy's Integral Theorem are satisfied.

Step 2: Verify Conditions for Cauchy's Integral Theorem

Cauchy's Integral Theorem states that if \(f(z)\) is analytic (holomorphic) within and on some closed contour \(C\), then \(\oint_{C} f(z) \, dz = 0\). We need to check if \(f(z) = z^3 - 1 + 3i\) is analytic inside and on the unit circle \(|z|=1\).

Step 3: Determine Analyticity of the Function

A polynomial function is entire, meaning it is analytic over the entire complex plane. Since \(f(z)\) is a polynomial, it is analytic everywhere, particularly on and within the unit circle \(|z|=1\).

Step 4: Apply Cauchy's Integral Theorem

Since the function \(f(z) = z^3 - 1 + 3i\) is analytic within and on the contour \(C\) (the unit circle), Cauchy's Integral Theorem applies. Therefore, \(\oint_{C} f(z) \, dz = 0\).

Step 5: Conclusion

Since all conditions for Cauchy's Integral Theorem are satisfied, the contour integral \(\oint_{C} f(z) \, dz\) evaluates to zero.

Key Concepts

Contour Integral

A contour integral is a way to integrate a complex-valued function along a specified path, or contour, in the complex plane. Think of it like tracing a path along a graph with a smooth pen stroke that begins and ends at the same point. The path is not necessarily straight and can curve, allowing us to explore more complex surfaces in the plane.

• The contour is a closed curve, meaning it loops back to the starting point.
• In mathematical notation, we often denote a contour integral with \(\oint_C f(z) \, dz \), where \(C\) represents the contour and \(f(z)\) is the function being integrated.
• To solve a contour integral, especially if involving a path along the unit circle, we need to ensure the function is analytic everywhere along, and inside, the contour.

In practical terms, certain theorems like Cauchy’s Integral Theorem simplify solving contour integrals by assuring us the integral equates to zero if the function meets specific criteria.

Analytic Function

An analytic function, also known as a holomorphic function, is a function that is smooth and differentiable at every point within a certain domain on the complex plane. This differentiability extends to both the function and all its derivatives. Imagine smoothly drawing a curve without any kinks or breaks; that's the level of smoothness and continuity required for analyticity.

• Analyticity means the function is not only differentiable but has derivatives of all orders in its domain.
• For a function like \(f(z) = z^3 - 1 + 3i\), being a polynomial, it is analytic over the entire complex plane.
• Cauchy’s Integral Theorem hinges on the property of analyticity inside a closed contour for setting the integral to zero, as analytic functions can't have sudden "jumps" or irregularities in their domain.

An intuitive way to understand this is to think of it like a perfectly calm lake surface: any tiny movement (infinitesimal derivative) on it is predictable and smoothly flows without disruption.

Complex Plane

The complex plane is a two-dimensional plane that allows us to visualize and work with complex numbers. If you imagine a regular Cartesian plane, the complex plane is quite similar. However, it integrates two key components of complex numbers: the real and the imaginary parts.

• The horizontal axis typically represents the real part of the complex number, \(Re(z)\).
• The vertical axis represents the imaginary part, \(Im(z)\).
• Each point on the complex plane corresponds to a complex number \(z = x + yi\), where \(x\) is the real part and \(y\) is the imaginary part.

In the context of contour integration, the complex plane serves as the canvas where contours (closed paths) are drawn, and functions are mapped. There's a certain beauty in this area of calculus and analysis, as it combines visual geometric interpretation with algebraic calculations to solve integrals.