Question

Verify that $$ \mathbf{y}_{1}=e^{\alpha t}(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) \quad \text { and } \quad \mathbf{y}_{2}=e^{\alpha t}(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) $$ are the real and imaginary parts of $$ e^{\alpha t}(\cos \beta t+i \sin \beta t)(\mathbf{u}+i \mathbf{v}) $$

Short answer

The given functions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are the real and imaginary parts, respectively, of the complex function \(e^{\alpha t}(\cos \beta t+i \sin \beta t)(\mathbf{u}+i \mathbf{v})\). The real part, \(\mathbf{y}_{1}\), is given by: $$ \mathbf{y}_{1}=e^{\alpha t}(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) $$ The imaginary part, \(\mathbf{y}_{2}\), is given by: $$ \mathbf{y}_{2}=e^{\alpha t}(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) $$

Step-by-step solution

Analyze the given complex function

Let's first analyze the given complex function, and rewrite it in a helpful form: $$ e^{\alpha t}(\cos \beta t+i \sin \beta t)(\mathbf{u}+i \mathbf{v}) = e^{\alpha t}[(\cos \beta t+i \sin \beta t)(\mathbf{u}+i \mathbf{v})] $$

Distribute the product

Next, we need to distribute the product of the complex expressions $$ (\cos \beta t+i \sin \beta t)(\mathbf{u}+i \mathbf{v}) = (\cos \beta t \cdot \mathbf{u} + i \cos \beta t \cdot \mathbf{v}) + (i \sin \beta t \cdot \mathbf{u} - \sin \beta t \cdot \mathbf{v}) $$

Simplify the expression

Now, group the real and imaginary parts together $$ (\cos \beta t \cdot \mathbf{u} - \sin \beta t \cdot \mathbf{v}) + i(\cos \beta t \cdot \mathbf{v} + \sin \beta t \cdot \mathbf{u}) $$

Distribute the exponential

Next, we distribute the exponential term \(e^{\alpha t}\) to both the real and imaginary parts $$ e^{\alpha t}(\mathbf{u} \cos \beta t - \mathbf{v} \sin \beta t) + i e^{\alpha t} (\mathbf{u} \sin \beta t + \mathbf{v} \cos \beta t) $$

Compare with given functions

Now, we can compare this expression with the given functions: Real Part: $$ \mathbf{y}_{1}=e^{\alpha t}(\mathbf{u} \cos \beta t-\mathbf{v} \sin \beta t) $$ Imaginary Part: $$ \mathbf{y}_{2}=e^{\alpha t}(\mathbf{u} \sin \beta t+\mathbf{v} \cos \beta t) $$ The real and imaginary parts of the complex function match the given functions \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\), respectively. Hence, we have successfully verified that the given functions are the real and imaginary parts of the given complex function.

Key concepts

Understanding Real and Imaginary Parts of Complex Functions

Complex functions are fundamental in differential equations, particularly when dealing with oscillatory systems. A complex number typically consists of a real part and an imaginary part, which can be represented as a combination of vectors on a complex plane.

When evaluating complex functions, especially in relation to solutions of differential equations, it's helpful to separate these parts to understand the behavior of the system. For instance, in the exercise provided, the real part of the complex function, \(\mathbf{y}_{1}\), relates to a damped or growing cosine function, while the imaginary part, \(\mathbf{y}_{2}\), corresponds to a sine function, each modified by the exponential term \(e^{\alpha t}\).

Separating a complex function into its real and imaginary components is crucial for analysis as it simplifies the process of interpretation, allowing us to utilize known strategies for solving real and imaginary differential equations.

Exponential Functions in Complex Analysis

Exponential functions play a vital role in complex analysis, particularly with the form \(e^{\alpha t}\), where \(\alpha\) is a complex constant and \(t\) represents time or another independent variable. In our exercise, the exponential function \(e^{\alpha t}\) provides the amplitude modulation for the trigonometric functions, which represent the oscillatory behavior of the solution to the differential equation.

The exponential can either grow or decay over time, depending upon the sign of the real part of \(\alpha\). If the real part of \(\alpha\) is positive, the function grows; if it's negative, it decays. This characteristic is critical for understanding the dynamics of systems modeled by differential equations, including signal processing, control systems, and physics.

Trigonometric Functions in Complex Analysis

Trigonometric functions arise frequently in complex analysis, particularly when dealing with periodic phenomena. They are the building blocks for representing oscillatory solutions in differential equations.

In our exercise, the trigonometric functions \(\cos\beta t\) and \(\sin\beta t\) describe oscillations with a frequency determined by \(\beta\). When paired with the exponential function, they capture both the frequency of oscillation and the rate of exponential growth or decay, determined by \(\alpha\).

This interplay between exponential and trigonometric functions can result in diverse behaviors from simple harmonic motion to damped or growing oscillations, which is why they are essential in the study of complex functions in differential equations.

Euler's Formula and Its Application

Euler's formula, given by \(e^{i\theta} = \cos\theta + i\sin\theta\), is a cornerstone in relating the trigonometric functions to exponential functions within complex analysis.

This elegant identity links the seemingly disparate worlds of exponentials and trigonometry and offers a powerful tool for simplifying the analysis of complex functions. In the context of the provided exercise, Euler's formula helps to verify that the expression \(e^{\alpha t}(\cos\beta t + i\sin\beta t)(\mathbf{u} + i\mathbf{v})\) indeed captures the real and imaginary parts identified as \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\).

When the exponential function includes a complex exponent, as in our verification process, Euler's formula allows us to convert the complex exponentials into a combination of simple, more manageable trigonometric functions, facilitating their analysis and interpretation in differential equations.