Question

Evaluate the following line integrals in the complex plane by direct integration, that is, as in Chapter \(6,\) Section \(8,\) not using theorems from this chapter. (If you see that a theorem applies, use it to check your result.) In finding complex Fourier series in Chapter \(7,\) we showed that $$\int_{0}^{2 \pi} e^{i n x} e^{-i m x} d x=0, \quad n \neq m$$ Show this by applying Cauchy's theorem to $$\oint_{C} z^{n-m-1} d z, \quad n > m$$ where \(C\) is the circle \(|z|=1\). (Note that although we take \(n > m\) to make \(z^{n-m-1}\) analytic at \(z=0,\) an identical proof using \(z^{m-n-1}\) with \(n < m\) completes the proof for all \(n \neq m .)\)

Short answer

The integral \( \int_{ 0}^ {2\pi} e^{i n x \cdot e^{-i m x } d x = 0 \) for \( n \eq m \). The proof uses Cauchy's theorem on \(\oint_{ 0}^{ 2\pi} z^{n-m-1} dx =0\)

Step-by-step solution

- Understand the given integral

Given the integral: \[ \int_{0}^{2 \pi} e^{i n x} e^{-i m x} d x \] The aim is to prove this integral is zero for \(n eq m\).

- Simplify the integral

Combine the exponents: \[ \int_{0}^{2 \pi} e^{i (n-m) x} d x \] Define \( k = n - m \) to give: \[ \int_{0}^{2 \pi} e^{i k x} d x \]

- Apply the Cauchy theorem

Evaluate the line integral: \[ \oint_{C} z^{n - m - 1} d z \] where \( C \) is the circle \( |z| = 1 \).

- Use the parametrization of the circle

Parametrize the contour \( C \) as \( z = e^{i x} \) with \( x \) ranging from \( 0 \) to \( 2\pi \). Thus, \( d z = i e^{i x} d x \), which yields: \[ \begin{aligned} &\text{Substitute } z = e^{i x} \text{ and } d z = i e^{i x} d x \text{ into the integral}, \ &\text{to get:} \ \text{\} \ \oint_{C} z^{k - 1} d z = \ \oint_{0}^{2 \pi} (e^{ix})^{k - 1} \cdot i e^{ix} dx = \iff \ i \cdot \oint_{0}^{2 \pi} carriedout (e^{ix})^{k } dx = \iff \ \ i \int_{0}^{2 \pi} e^{i (k) x} d x \end{aligned} \]

- Evaluate the integral

The integral can be written as: \[ i \int_{0}^{2 \pi} e^{i k x} d x \] Recall that for nonzero integer values of \( k \), \( \int_{0}^{2 \pi} carriedout e^{i k x} dx = 0 \), so: \[ i \cdot 0 = 0 \]

- Conclusion

From the computation above, verify that the integral is zero when \( n \eq m \):. \(\ \int_{ 0}^ {2\pi} e^{i n x \cdot e^{-i m x } d x = \ 0 \)

Key concepts

Cauchy's theorem

Cauchy's theorem is a fundamental result in complex analysis. It states that for any closed contour \(C\) and any function \(f(z)\) that is analytic (holomorphic) on and inside \(C\), the integral of \(f(z)\) over the contour \(C\) is zero. In mathematical form, this is:
\[ \oint_{C} f(z) \,dz = 0 \]
This theorem helps simplify complex integration significantly. In the given problem, we use Cauchy's theorem to show that the integral of \(z^{n-m-1}\) around the unit circle is zero when \(n eq m\). Since \(z^{n-m-1}\) is analytic based on the condition given (\(n > m\)), the theorem applies as such.

Complex Fourier series

A complex Fourier series is a way to represent a periodic function as a sum of complex exponentials. It is particularly useful in solving differential equations and signal processing. The series has the form:
\[ f(x) = \sum_{n=-\infty}^{+\infty} c_n e^{inx} \]
Here, \(c_n\) are the Fourier coefficients, calculated by:
\[ c_n = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) e^{-inx} \, dx \]
In the exercise, we encounter an integral of the form \(\int_{0}^{2 \pi} e^{inx} e^{-imx} dx = 0\), for \(n \eq m\). This stems from the orthogonality property of the complex exponentials, where such integrals result in zero unless \(n = m\).

Line integrals

In complex analysis, a line integral is an integral where the function to be integrated is evaluated along a contour in the complex plane. It’s similar to integrating a function along a path in real calculus, but in the complex plane, we deal with paths in \(\mathbb{C}\). For a function \(f(z)\) and a contour \(C\), the line integral is denoted:
\[ \oint_{C} f(z) \, dz \]
In the example, we convert the given function in terms of the parameter \(z\). The parametrization of the circle \(|z| = 1\) as \(z = e^{ix}\) simplifies the integral, making it easier to evaluate. This change of variable harnesses the periodic properties of the complex exponential function.

Parametrization in complex analysis

Parametrization involves expressing a contour in terms of a parameter. For instance, for a circle with radius 1, a common parametrization is \(z(t) = e^{it}\) where \(t\) ranges from \(0\) to \(2\pi\). This allows us to convert complex integrals into a more familiar form of real integrals.

In the given exercise, we parametrize the contour \(C\) as \(z = e^{ix}\) with \(x\) ranging from \(0\) to \(2\pi\). This helps us transform the integral as:
\[ dz = ie^{ix} dx \]
Substituting these, the integral simplifies to evaluating \(\int_{0}^{2 \pi} e^{i (k)x} dx\), which can be computed using standard integration techniques.