Question

Use Cauchy's theorem or integral formula to evaluate the integrals. \(\oint \frac{e^{3 z} d z}{z-\ln 2}\) if \(C\) is the square with vertices \(\pm 1 \pm i\)

Short answer

The value of the integral is 8.

Step-by-step solution

- Understand the Integral and Contour

Identify the function inside the integral and the contour along which integration is performed. The function given is \(\frac{e^{3z}}{z-\text{ln}2}\) and the contour \(C\) is the square with vertices \( \pm 1 \pm i \).

- Identify the Singularity

Recognize that the integrand has a singularity at \( z = \text{ln}2 \). Verify if this singularity lies inside the contour \(C\). Since \( \text{ln}2 \) is approximately \(0.69\) which lies within the square with vertices \( \pm 1 \pm i \), the singularity is indeed inside the contour.

- Apply Cauchy's Integral Formula

Use Cauchy's Integral Formula: \( \frac{1}{2\text{π}i} \oint_{C} \frac{f(z)}{z - z_0} dz = f(z_0) \) for a function \( f(z)\). Here, our \( f(z) = e^{3z} \) and \( z_0 = \text{ln}2\).

- Evaluate the Integral

By Cauchy's Integral Formula, we have: \(\frac{1}{2\text{π}i} \oint_{C} \frac{e^{3z}}{z - \text{ln}2} dz = e^{3 \text{ln}2}\). Thus, \(\frac{1}{2\text{π}i} \oint_{C} \frac{e^{3z}}{z - \text{ln}2} dz \) simplifies to \( e^{3 \text{ln} 2} \). Converting \( e^{3 \text{ln}2}\) to exponential terms, we get \( 2^3 = 8 \). Therefore, \(\frac{1}{2\text{π}i} \oint_{C} \frac{e^{3z}}{z - \text{ln}2} dz = 8\).

Key concepts

singularity

In complex analysis, a singularity refers to a point where a function is not defined or does not behave nicely. For example, in the function \(\frac{e^{3z}}{z - \text{ln}2}\), the point \(z = \text{ln}2\) is a singularity because substituting \(z = \text{ln}2\) would make the denominator zero, resulting in an undefined expression.
Singularities can be classified into different types such as:

• Removable Singularities: Points where the function can be redefined to remove the undefined behavior.
• Poles: Points where the function's magnitude becomes infinite.
• Essential Singularities: Points where the function's behavior is chaotic and does not settle into any pattern.

In this problem, \(z = \text{ln}2\) is a simple pole, which is a type of singularity where the function behaves like \(\frac{1}{(z-\text{ln}2)}\) near the point.

contour integration

Contour integration is a key technique in complex analysis that involves integrating a complex function around a specific path or 'contour' in the complex plane. The contour can be a circle, a square, or any closed loop.
When evaluating integrals using contour integration, it's important to:

• Identify the function to be integrated.
• Determine the path (contour) along which the integration will occur.
• Check if the singularities of the function lie within the contour.

In this exercise, the contour \(C\) is a square with vertices at \(\pm 1 \pm i\). We envision tracing this square in a counterclockwise direction, enclosing the singularity \(\text{ln}2\). The contour helps us apply powerful theorems like Cauchy's Integral Theorem or Cauchy's Integral Formula to evaluate integrals around singularities.

Contour integration simplifies the evaluation of complex integrals dramatically, especially when dealing with functions having singularities.

Cauchy's Integral Formula

Cauchy's Integral Formula is a cornerstone of complex analysis which provides the value of integrals for functions that are analytic inside and on some closed contour that encloses singularities. It states:
\[ \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z - z_0} dz = f(z_0) \]
This formula implies that if you know a function \(f(z)\) is analytic inside and on a closed contour \(C\), you can compute the value at any point \(z_0\) inside \(C\) using the integral around the contour.

In the provided problem, we have:

• \(f(z) = e^{3z}\)
• \(z_0 = \text{ln}2\)

Using Cauchy's Integral Formula, we substitute these values:
\[ \frac{1}{2\pi i} \oint_{C} \frac{e^{3z}}{z - \text{ln}2} dz = e^{3 \text{ln} 2} = e^{3 \cdot \text{ln} 2} = (e^{\text{ln} 2})^3 = 2^3 = 8 \]
Hence, the integral \( \oint_{C} \frac{e^{3z}}{z - \text{ln}2} dz \) evaluates to \(8 \cdot 2\pi i\).

Cauchy's Integral Formula simplifies complex contour integrals by transforming them into evaluations of simpler functions at specific points.