Question

$${\int }_{2\pi }^{2\pi +ix}{e}^{2iz}dz$$

Short answer

$\frac{i}{2}(1-e^{-2x})$

Step-by-step solution

- Understand the Path of Integration

The integral is taken along a path from $2\pi$ to $2\pi +ix$ in the complex plane. The path is a straight line parallel to the imaginary axis.

- Parametrize the Path

Parametrize the path $z(t)=2\pi +it$ where $t$ goes from $0$ to $x$. Therefore, $dz=idt$.

- Substitute the Parameterization into the Integral

Substitute $z(t)=2\pi +it$ and $dz=idt$ into the integral: $${\int }_{2\pi }^{2\pi +ix}{e}^{2iz}dz={\int }_0^x{e}^{2i(2\pi +it)}idt.$$

- Simplify the Integral Expression

Simplify the integrand: $$e^{2i(2\pi +it)}=e^{4i\pi }{e}^{-2t}=1\cdot e^{-2t}=e^{-2t}.$$ Hence the integral becomes: $${\int }_0^{x}i{e}^{-2t}dt.$$

- Integrate

Integrate $i{e}^{-2t}$ with respect to $t$: $${\int }_0^{x}i{e}^{-2t}dt=i{\left[\frac{e^{-2t}}{-2}\right]}_0^x=-\frac{i}{2}[e^{-2x}-1].$$

- Simplify the Result

Simplify the final expression: $$-\frac{i}{2}[e^{-2x}-1]=\frac{i}{2}(1-e^{-2x}).$$

Key concepts

Complex Plane

The complex plane is a two-dimensional plane used to visualize complex numbers. Each point on this plane represents a complex number. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. In this exercise, we integrate along a path in the complex plane. The path starts at a point on the real axis, specifically at the point corresponding to the complex number $2\pi$, and ends at the point $2\pi +ix$. This integration path is a straight line parallel to the imaginary axis, indicating a vertical movement from $2\pi$ upward by a distance $x$ along the imaginary axis.

Parameterization

Parameterization is a method for describing a path in the complex plane by expressing it as a function of a single variable, usually $t$. This helps simplify complex integrals by converting them into terms of a single variable. Here, the path from $2\pi$ to $2\pi +ix$ is parameterized by letting $z(t)=2\pi +it$, where $t$ varies from 0 to $x$. This means we substitute $z$ with $z(t)$ and $dz$ with $idt$ in the integral, transforming our original integral into one with limits from $t=0$ to $t=x$.

Exponential Function

The exponential function $e^{2iz}$ appears in the integrand of our complex integral. Complex exponential functions have properties that are helpful in integration. By Euler's formula, $e^{ix}=\cos (x)+i\sin (x)$. For our parameterized path, we substitute $z=2\pi +it$, giving us $e^{2i(2\pi +it)}$. Using properties of exponents, this simplifies to $e^{4i\pi }{e}^{-2t}$. Since $e^{4i\pi }=1$ (as $4\pi$ is a multiple of $2\pi$), the function further reduces to $e^{-2t}$. This simplification transforms our integral into a more manageable form.

Path of Integration

The path of integration is crucial in evaluating complex integrals. In our exercise, we move along a straight path parallel to the imaginary axis. Understanding this path is necessary for parametrizing the integral and correctly substituting in the exponential part. Once we parametrize the path and integrate over the new variable, the integral becomes easier to handle. Integrating along complex paths often simplifies to integrals involving real values, as seen when we ended with an integral of $i{e}^{-2t}$ from 0 to $x$. This straightforward path helps minimize complexity, making the problem solvable step-by-step.