Question

If \(C\) is a circle of radius \(\rho\) about \(z_{0}\), show that $$ \oint_{c} \frac{d z}{\left(z-z_{0}\right)^{n}}=2 \pi i \quad \text { if } n=1 $$ but for any other integral value of \(n\), positive or negative, the integral is zero. Hint: Use the fact that \(z=z_{0}+\rho e^{1 \theta}\) on \(C\).

Short answer

\(\oint_{C} \frac{dz}{(z-z_{0})^n}\) is \2i\ for \(n=1\) and zero for other integral \(n\).

Step-by-step solution

- Understand the Problem

Given a circle of radius \(\rho\) centered at \(z_{0}\), and the integral to be evaluated along this circle. Our goal is to evaluate \(\frac{dz}{(z-z_{0})^{n}}\) for \(n=1\) and for other integral values of \(n\).

- Parameterize the Circle

Use the hint that \(z = z_{0} + \rho e^{i \theta}\), where \(\theta\) ranges from 0 to 2\(p\). This represents any point on the circle.

- Differentiate with Respect to \(\theta\)

Find \(dz\). Since \(z = z_{0} + \rho e^{i \theta}\), differentiating gives us \(dz = i\rho e^{i\theta} d\theta\).

- Substitute \(z\) and \(dz\)

Substitute \(z = z_{0} + \rho e^{i \theta}\) and \(dz = i\rho e^{i\theta} d\theta\) into the given integral \(\frac{dz}{(z-z_{0})^n}\). We get \(\frac{i\rho e^{i\theta} d\theta}{(\rho e^{i\theta})^n} \).

- Simplify the Integral

Simplify the integral: \(\frac{i\rho e^{i\theta} d\theta}{(\rho e^{i\theta})^n} = \frac{i\rho e^{i\theta} d\theta}{\rho^n e^{in\theta}} = \frac{i d\theta}{\rho^{n-1} e^{i(n-1)\theta}} \).

- Evaluate for \(n=1\)

When \(n=1\), the integral becomes \(\frac{i d\theta}{\rho^{0} e^{i(1-1)\theta}} = i \oint_{C} d\theta = i (2p) = 2i \).

- Evaluate for \(n eq 1\)

For any other integral value of \(n\), \(n eq 1\), the integral becomes \(\frac{i d\theta}{\rho^{n-1} e^{i(n-1)\theta}}.\) Integration of \(e^{i(n-1)\theta}\) from 0 to 2\(p\) results in zero due to periodicity of the exponential function.

Conclusion

Thus, \(\oint_{C} \frac{dz}{(z-z_{0})^n} = 2i \) when \(n=1\), and zero for any other integral value of \(n\).

Key concepts

Contour Integration

Contour integration is a method used in complex analysis, where we integrate a complex function along a certain path (or contour) in the complex plane. This technique is particularly powerful as it allows the evaluation of integrals that would be difficult or impossible to solve using standard calculus methods.

In our problem, we are working with the contour integral \(\oint_{C} \frac{d z}{(z-z_{0})^n}\), where C is a circle of radius \(\rho\) centered at \(z_{0}\).
This type of integral is often simplified by parameterizing the contour, letting us transform the complex integral into a more standard form.

By understanding and applying contour integration, we can solve a wide variety of problems in complex analysis, including those that appear in calculus and differential equations.

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a fundamental result in complex analysis which states that the integral of a holomorphic (complex differentiable) function over a closed contour is zero, given the function has no singularities inside the contour.

This theorem simplifies the evaluation of complex integrals significantly. It allows us to understand why certain integrals evaluate to zero when conditions are met.

In our problem, for \(n eq 1\), our function \(\frac{d z}{(z-z_{0})^n}\) can be seen as non-holomorphic because it has a pole of order \(n\) at \(z = z_{0}\). Thus, Cauchy's Integral Theorem helps us conclude that for \(n eq 1\), the integral around the contour is zero. For \(n = 1\), the function has a simple pole, aligning directly with the conditions where Cauchy's Integral Formula applies.

Parameterization of Curves

Parameterization is a method where we describe a curve using a parameter, typically denoted as \( \theta \) or \( t \).

In our problem, we parameterize the circle of radius \( \rho \) centered at \( z_0 \) by letting \(\ z = z_0 + \rho e^{i\theta}\).

This allows us to express any point on the circle in terms of the parameter \( \theta \) varying from 0 to 2\ \pi. By differentiating this parameterization, we can also find the differential element \( dz = i\rho e^{i\theta} d\theta \).

This transformation simplifies the integral dramatically, letting us rewrite the complex integral in terms of an easily solvable real integral.

Parameterization of curves is a cornerstone technique in contour integration, aiding in transitioning from complex forms to manageable real-number forms.

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis for evaluating contour integrals. It states that if a function has isolated singularities inside a contour, the integral around the contour is 2πi times the sum of residues at those singularities.

In simpler terms, the residue of a function at a singularity is a coefficient representing the behavior of the function near that point.

In our exercise, when \(n = 1\), our function \(\frac{1}{(z-z_{0})}\) has a simple pole at \(z = z_0\). The residue at this point is 1. Thus, by the Residue Theorem, the value of the contour integral around this pole is 2πi times the residue, leading to \( 2\pi\i\).

For \(n eq 1\), the residues are zero because the integrand's form does not contribute non-zero values.

Hence, understanding residues helps us solve complex integrals efficiently, identifying contributions from essential singularities quickly.