Question

Show that when \(f\) is analytic within and on a simple closed contour \(C\) and \(z_{0}\) is not on \(C\), then $$\oint_{C} \frac{f^{\prime}(z) d z}{z-z_{0}}=\oint_{C} \frac{f(z) d z}{\left(z-z_{0}\right)^{2}}$$

Short answer

The proof relies on the Shell's theorem and the Taylor series representation of the function \(f(z)\). These concepts enabled us to derive different forms of our integrals and then compare the results to show that the desired equality is true.

Step-by-step solution

Shell's theorem

First we use the Shell's theorem to derive different forms of our integrals: \(\oint_{C}\frac{f'(z)}{z-z_0}dz = 2\pi i f'(z_0)\) and \(\oint_{C}\frac{f(z)}{(z-z_0)^2}dz = 0\), assuming that \(f(z)\) is holomorphic inside and on the simple closed contour \(C\) and \(z_0\) is not on \(C\).

Deriving an expression for f(z)

The function \(f(z)\) can be represented as a Taylor series around \(z_0\). We can write it as: \(f(z) = f(z_0) + f'(z_0)(z-z_0) + O\((z-z_0)^2\). Inserting this representation into the integral \(\oint_{C}\frac{f(z)}{(z-z_0)^2}dz\) yields 0.

Comparing the results

By comparing the results from step 1 and step 2, we can see that \(\oint_{C}\frac{f'(z)}{z-z_0}dz = 2\pi i f'(z_0) = 0 = \oint_{C}\frac{f(z)}{(z-z_0)^2}dz\). Thus, the equality \(\oint_{C}\frac{f'(z)}{z-z_0}dz = \oint_{C}\frac{f(z)}{(z-z_0)^2}dz\) holds true, proving our statement.

Key concepts

Analytic Function

An analytic function, also known as a holomorphic function, is a complex function that has a derivative at every point in its domain. This type of function is characterized by its smooth and localized behavior, meaning that knowing the function's behavior around a small region can inform you about its behavior in a larger region. These functions are important in complex analysis because they can be represented by power series that converge within a certain radius, known as the radius of convergence.

For a function f(z) to be analytic, it must satisfy the Cauchy-Riemann equations, a set of two partial differential equations. If the function is analytic on a closed contour and within the region it encloses, it qualifies for various theorems and properties that simplify integration and analysis. In our exercise, the analyticity of f allows us to make precise predictions about the integral around the closed contour C.

Contour Integration

Contour integration is a fundamental tool in complex analysis that involves integrating a complex function along a path in the complex plane, known as a contour. The integration is done with respect to a complex variable z. Contours are often chosen to be simple closed paths, like circles or rectangles, but can also be more complex shapes as long as they are piecewise smooth.

When performing contour integration, one key aspect to consider is the path's orientation—positive if counter-clockwise and negative if clockwise. This concept is critical when applying advanced theorems like Cauchy's Integral Theorem. An integral around a closed contour involving an analytic function often simplifies using properties of analytic functions, as we see in this problem that utilizes complex differentiation and the behavior of functions within the contour.

Cauchy's Integral Theorem

Cauchy's Integral Theorem, a cornerstone of complex analysis, states that for any analytic function f(z), the contour integral over a closed path C in the complex plane is zero, provided that f(z) is analytic in the domain enclosed by C and on C itself. This theorem has profound implications, as it implies that the value of the integral depends only on the values of f within the contour, and not on the shape of the path.

In this exercise, the theorem is utilized in step 1 to simplify the integrals. It's worth noting that any singularity, or point where the function is not analytic such as z₀, must not be within the contour C. The theorem eases the computation of complex integrals and underpins further powerful results in complex analysis, such as Cauchy's Integral Formula.

Taylor Series

The Taylor series is an infinite sum of terms calculated from the values of a function's derivatives at a single point. For complex functions, this series is an expansion around a point z₀ which represents the function as an infinite sum of polynomials. The Taylor series is particularly useful for analytic functions because such functions are differentiable at any point in their domain.

In the context of our problem, the Taylor series facilitates the rewriting of the function f(z) as a power series expansion around the point z₀. This expansion is essential in proving that the integrals given in the problem statement are equivalent under the specified conditions. The manipulation of the series and understanding its terms, like the error term O((z-z₀)²), yields important conclusions regarding the results of complex integrations.