Problem 18 - Complex Analysis for Mathematics and Engineering

Question

Let \(f\) be analytic in the simply connected domain \(D\) and let \(z_{1}\) and \(z_{2}\) be two complex numbers that lie interior to the simple closed contour \(C\) having positive orientation that lies in \(D\). Show that, \(\frac{f\left(z_{2}\right)-f\left(z_{1}\right)}{z_{2}-z_{1}}=\frac{1}{2 \pi i} \int_{C} \frac{f(z) d z}{\left(z-z_{1}\right)\left(z-z_{2}\right)}\) State what happens when \(z_{2} \rightarrow z_{1}\).

Step-by-Step Solution

Answer

The given expression holds, and as \(z_2\) approaches \(z_1\), it equals \(f'(z_1)\).

Step 1: Understanding the Problem

We need to show that the expression \(\frac{f(z_2) - f(z_1)}{z_2 - z_1}\) can be expressed as a contour integral involving the analytic function \(f\) over a contour \(C\) that encloses the points \(z_1\) and \(z_2\). Then examine the behavior of this expression as \(z_2\) approaches \(z_1\).

Step 2: Setting Up the Integral Expression

Recall from complex analysis that if \(f\) is analytic in domain \(D\), the integral \(\int_C \frac{f(z)}{(z-z_1) (z-z_2)} dz\) is valid within \(C\) if neither \(z_1\) nor \(z_2\) is on \(C\), and \(C\) encloses \(z_1\) and \(z_2\).

Step 3: Expressing the Derivative as a Limit

Use the Cauchy integral formula for \(f(z)\): \[f(w) = \frac{1}{2\pi i} \int_{C} \frac{f(z) dz}{z-w}\],and apply it for both \(z_1\) and \(z_2\). From the definition of the derivative, we can express \[f'(w) = \lim_{z_2 \to z_1} \left( \frac{f(z_2) - f(z_1)}{z_2 - z_1} \right).\]

Step 4: Applying Cauchy's Integral Formula

From Cauchy's Integral Formula we can write: \[f(z_{1}) = \frac{1}{2 \pi i} \int_{C} \frac{f(z) dz}{z-z_1}\] and \[f(z_{2}) = \frac{1}{2 \pi i} \int_{C} \frac{f(z) dz}{z-z_2}\]. Subtract these two expressions and divide by \(z_2 - z_1\) to achieve the desired result.

Step 5: Simplifying the Expression

Using the linearity of integration and the common denominator, combine the two integrals: \[\frac{1}{2 \pi i} \int_{C} \frac{f(z)}{(z-z_1)(z-z_2)} dz\]which matches the right side of the original equation. Thus, the original claim is verified.

Step 6: Analyzing When \(z_2 \rightarrow z_1\)

As \(z_2\) approaches \(z_1\), the expression \(\frac{f(z_{2})-f(z_{1})}{z_{2}-z_{1}}\) tends to the derivative \(f'(z_1)\). This is from the definition of the derivative in complex analysis.

Key Concepts

Contour Integration

Contour integration plays a vital role in complex analysis. It's all about integrating a complex function over a path or contour in the complex plane. Imagine you are drawing a path on a piece of paper, and this path is your contour.
To perform contour integration, you need a function that is smooth (or analytic) and a path that forms a closed loop. The closed contour can be simple like a circle or more complex.
While doing contour integration, it's crucial to pay attention to the orientation of the contour. Usually, the path has a positive orientation, meaning that if you keep walking along it, the region you are enclosing is on your left-hand side.

• The integral of a function over a contour is deeply influenced by the function's behavior inside this closed path.
• If your function is analytic (no sudden jumps or holes) inside and on the contour, some mighty powerful theorems come into play!
• The process often simplifies considerably thanks to Cauchy's Theorems.

Understanding contour integration is crucial for grasping how complex functions interact when seen as 'flowing' around a loop in their domain.

Cauchy's Integral Formula

Cauchy's Integral Formula is a cornerstone of complex analysis. It shows the power of contour integration when dealing with analytic functions.
The formula states that if a function is analytic and enclosed by a closed contour, then all information about that function inside the contour is captured by the integral over its boundary. In math terms:
Let \( w \) be inside a contour \( C \), and \( f \) analytic on and inside \( C \):\[f(w) = \frac{1}{2\pi i} \int_{C} \frac{f(z) dz}{z-w}\]
This essentially allows you to compute the value of the function inside the area encircled by \( C \) via its surrounding values!
Here's why it's magical:

• It provides a direct connection between values of an analytic function inside a domain and its values on the boundary.
• The formula is immensely useful for calculating derivatives.
• If the derivatives exist at every point, the function is infinitely differentiable. These derivatives can also be computed using Cauchy's integral formulas for derivatives.

Using Cauchy's Integral Formula makes analyzing complex analytic functions much more manageable!

Analytic Functions

Analytic functions are central players in complex analysis. An analytic function can be seen as the complex equivalent of something more familiar—smooth real functions.
For a function to be analytic, it must be complex differentiable in a region. That means at every point in its domain, the function has a well-defined derivative.
What makes analytic functions so interesting?

• They're locally represented by a Taylor series. Around any point of analyticity, the function looks like an infinite polynomial.
• If it’s analytic everywhere in a region, tiny nuances and details of function behavior are fully captured by its series.
• Despite being delicate, they surprisingly endure small perturbations and remain analytic.

Most importantly, analytic functions adhere to the conditions needed for applying wonderful things like Cauchy's theorems and contour integrals.
Their predictability makes them incredibly useful for solving practical problems in physics and engineering! Understanding these features allows you to unlock the hidden potential of analytic functions in complex analysis.