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.) If $f(z)$ is analytic in the disk $|z|\le 2,$ evaluate ${\int }_0^{2\pi }{e}^{2i\theta }f\left(e^{i\theta }\right)d\theta$

Short answer

The integral equals $0$.

Step-by-step solution

Parameterize the given curve

Parameterize the unit circle in the complex plane using the substitution $$z=e^{i\theta }$$ where $\theta$ ranges from $0$ to $2\pi$.

Rewrite the integral

Substitute the parameterized form into the given integral: $${\int }_0^{2\pi }{e}^{2i\theta }f\left(e^{i\theta }\right)d\theta$$ becomes $${\int }_0^{2\pi }{e}^{2i\theta }f(z)\left(\frac{dz}{iz}\right)$$ where $dz=i{e}^{i\theta }d\theta$.

Simplify the integral

Substitute back the expression for $dz$: $$dz=i{e}^{i\theta }d\theta$$ So, $$d\theta =\frac{dz}{i{e}^{i\theta }}$$ The integral becomes: $${\int }_0^{2\pi }{e}^{2i\theta }f\left(e^{i\theta }\right)\left(\frac{dz}{i{e}^{i\theta }}\right)$$ Simplifying gives: $${\int }_0^{2\pi }{e}^{i\theta }f\left(e^{i\theta }\right)\frac{dz}{i}$$ $${\int }_{|z|=1}{e}^{i\theta }f(z)\frac{dz}{iz}$$

Integrate

Since $e^{i\theta }=z$ on the unit circle, the integral simplifies to: $$\frac{1}{i}{\int }_{|z|=1}f(z)dz$$ Since $f(z)$ is analytic inside $|z|\le 2$, by Cauchy's theorem, this integral around a closed contour where the function is analytic is equal to zero.

Conclude the solution

Therefore, $${\int }_0^{2\pi }{e}^{2i\theta }f\left(e^{i\theta }\right)d\theta =0$$

Key concepts

Line Integrals in the Complex Plane

A line integral in the complex plane allows us to integrate complex-valued functions over curves. Instead of integrating over intervals like in real calculus, we integrate over paths or contours. These paths are often parameterized curves.

To define a line integral, we use parameterization. For example, if we integrate a complex function along a unit circle, we can parameterize it using $z=e^{i\theta }$. This transformation converts our path into a form we can easily work with.

Understanding line integrals is crucial for more advanced concepts like complex integration and contour integration. They are foundational in evaluating integrals of analytic functions and applying theorems like Cauchy's theorem.

Analytic Functions

An analytic function is a complex function that is differentiable at every point within a region in the complex plane. These functions are also called holomorphic functions.

Analytic functions have several important properties:

• They have derivatives of all orders.
• They can be represented by a convergent power series within their radius of convergence.
• They satisfy the Cauchy-Riemann equations.

In the given exercise, $f(z)$ is stated to be analytic inside $|z|\le 2$, ensuring that it is smooth and continuous in that region.

Cauchy's Theorem

Cauchy's theorem is a fundamental theorem in complex analysis. It states that if $f(z)$ is analytic and its derivative is continuous within and on some simple closed contour $C$, then:

$${\int }_{C}f(z)dz=0$$

This theorem simplifies complex integrations significantly. In the exercise, after parameterization and simplification, the integral is evaluated on a unit circle, which is a closed contour. Since $f(z)$ is analytic in $|z|\le 2$, Cauchy's theorem tells us that the integral is zero.

Unit Circle Parameterization

Unit circle parameterization is a technique used to simplify complex integrals involving closed contours. The unit circle is parameterized by:

$$z=e^{i\theta }$$

where $\theta$ ranges from $0$ to $2\pi$. This transformation helps convert the integral into a more manageable form using complex exponentials.

In the exercise, we parameterize the unit circle to rewrite the integral. The parameterization simplifies otherwise complex-looking integrals and allows us to directly apply theorems like Cauchy's theorem.