Question

Prove rule \(\mathrm{C}\) by using (3.9). Hints: If \(f(z)\) has a pole of order \(n\) at \(z=a\), then \(f(z)=g(z) /(z-a)^{n}\) with \(g(z)\) analytic at \(z=a .\) By \((3.9)\) $$ \int_{c} \frac{g(z)}{(z-a)} d z=2 \pi i g(a) $$ with \(C\) a contour inclosing \(a\) but no other singularities. Differentiate this equation \((n-1)\) times with respect to \(a\). (Or, use Problem \(3.21 .\) )

Short answer

By differentiating the given integral formula \((n-1)\) times with respect to \(a\), and applying Leibniz's rule, we obtain the given rule.

Step-by-step solution

Express the function using the given pole

Given that if a function \(f(z)\) has a pole of order \(n\) at \(z=a\), we can write \(f(z)\) as \(f(z) = \frac{g(z)}{(z-a)^{n}}\), where \(g(z)\) is analytic at \(z=a\).

Apply integral formula

Using equation (3.9), which states that \( \int_{C} \frac{g(z)}{(z-a)} dz = 2\pi i g(a)\) for a contour \(C\) enclosing \(a\) but no other singularities, we incorporate our function \(f(z)\).

Modify the integral

We now consider the integral \(\int_{C} f(z) dz = \int_{C} \frac{g(z)}{(z-a)^{n}} dz\). To proceed, we need to differentiate this integral \((n-1)\) times with respect to \(a\).

Differentiate with respect to \(a\)

Differentiate both sides of the equation \(\int_{C} \frac{g(z)}{(z-a)} dz = 2\pi i g(a)\) \((n-1)\) times with respect to \(a\). This gives us: $$ \frac{d^{n-1}}{da^{n-1}} \left(\int_{C} \frac{g(z)}{(z-a)} dz \right) = 2\pi i \frac{d^{n-1}}{da^{n-1}} g(a). $$

Use Leibniz's Rule for differentiation under the integral sign

Applying Leibniz’s rule for differentiation under the integral sign, and recalling that \(g(z)\) is analytic (hence all of its derivatives exist and are also analytic), we get: $$ \int_{C} \frac{(n-1)! g(z)}{(z-a)^{n}} dz = 2\pi i \frac{d^{n-1}}{da^{n-1}} g(a). $$

Simplify

Thus, recognizing that the integral we started with represents \(f(z)\) and integrating/simplifying accordingly, results in: $$ \int_{C} f(z) dz = 2\pi i (n-1)! \frac{d^{n-1}}{da^{n-1}} g(a). $$

Key concepts

Pole of Order n

In complex analysis, a 'pole of order n' refers to a specific type of singularity that a function might have at a certain point, say z = a. When we say a function f(z) has a pole of order n at z = a, it means that the function can be expressed as:

$$ f(z) = \frac{g(z)}{(z-a)^n} $$

Here, g(z) is an analytic function at z = a, which means g(z) is well-behaved and does not have any singularity at z = a.

The integer n indicates that the denominator (z - a) is raised to the power n, which dictates how 'severe' the singularity or pole is. For example, a pole of order 2 at z = a implies (z - a)^2 in the denominator.

Understanding poles is crucial because they help us analyze and evaluate functions that might seem complex at first glance. By breaking them down into their components, we can more easily manipulate and understand the underlying function.

Analytic Function

An analytic function, sometimes known as a holomorphic function, is a function that is complex differentiable in a neighborhood of every point in its domain. If a function g(z) is analytic at z = a, it means:

• g(z) is differentiable at z = a.
• It can be expressed as a convergent power series in terms of (z - a) in the neighborhood around a.

For the function g(z), its analyticity ensures that its derivatives of all orders exist and are continuous. This property is pivotal when considering poles and contour integration, as it essentially guarantees that the function behaves nicely around the point z = a except for potential singularities.

In our original exercise, recognizing g(z) as analytic allowed us to use it comfortably within the integral and apply calculus principles to differentiate and evaluate it.

Contour Integration

Contour integration involves integrating a complex function along a path, or contour, in the complex plane. The idea is to evaluate the integral of a complex function over a specified path, such as a circle or any other curve, which encloses points of interest.

In our context, we were given:

$$ \int_{C} \frac{g(z)}{(z-a)} dz = 2\pi i g(a) $$

Here, the contour C encloses the point a but not any other singularities. This is a standard result in complex analysis often derived from the residue theorem.

When we integrate a function around a point where it has a singularity or pole, essential characteristics like the nature of the pole and the function's behavior near the pole can be captured through this integration technique. Contour integration is a powerful tool for evaluating many complex integrals, especially when regular methods struggle.

In our exercise, we utilized contour integration to manage and simplify the integral involving our function f(z) with its pole.

Leibniz's Rule

Leibniz's Rule refers to a formula for the differentiation of an integral with respect to a parameter. In simpler terms, if we have an integral dependent on a variable and we want to differentiate under the integral sign, Leibniz's rule tells us how to do it correctly.

For a function and integral F(t)\right):

$$ F(t) = \int_{a}^{b} f(x,t) dx $$

Leibniz's rule states:

$$ \frac{d}{dt} F(t) = \int_{a}^{b} \frac{\partial f(x,t)}{\partial t} dx $$

In our exercise, we used a similar principle to differentiate an integral multiple times to solve our function involving the pole. By recognizing that derivatives can be taken under the integral sign, we could simplify and manage our expressions clearly.

This rule allowed us to systematically differentiate (n-1) times, making our complex integration in terms of a parameter straightforward and manageable.