Problem 36 In Problems \(33-36\), find all ... [FREE SOLUTION]

Chapter 4: Problem 36

In Problems \(33-36\), find all complex values of \(z\) satisfying the given equation. \(e^{2 z}+e^{z}+1=0\)

Short Answer

Expert verified

The solutions for \( z \) are \( z = i \left( \frac{2\pi}{3} + 2k\pi \right) \) or \( z = i \left( \frac{4\pi}{3} + 2k\pi \right) \), for integer \( k \).

Step by step solution

Write the Equation in Terms of a Single Variable

Let us set \( w = e^z \). Then the given equation \( e^{2z} + e^z + 1 = 0 \) can be rewritten as a quadratic equation in terms of \( w \): \( w^2 + w + 1 = 0 \).

Solve the Quadratic Equation

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( w \), where \( a = 1, b = 1, c = 1 \). This gives \( w = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \). Hence, \( w = \omega, \omega^2 \), where \( \omega = \frac{-1 + i\sqrt{3}}{2} \) is a cube root of unity and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

Find Values of \( z \) Using \( w = e^z \)

Since \( w = e^z \), we need to find \( z \) such that \( e^z = \omega \) or \( e^z = \omega^2 \). Using \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), we have \( \omega = e^{i\frac{2\pi}{3}} \) and \( \omega^2 = e^{i\frac{4\pi}{3}} \). Therefore, the solutions for \( z \) are \( z = i \left( \frac{2\pi}{3} + 2k\pi \right) \) and \( z = i \left( \frac{4\pi}{3} + 2k\pi \right) \), where \( k \) is any integer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Numbers

Complex numbers extend the idea of one-dimensional real numbers into the two-dimensional complex plane using the imaginary unit, represented as \( i \), where \( i^2 = -1 \). A complex number is expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers. Here, \( a \) is the real part and \( b \) is the imaginary part.

Imaginary Unit \( i \): This unit satisfies the property \( i^2 = -1 \).
Real Part: The \( a \) in \( a + bi \).
Imaginary Part: The \( b \) in \( a + bi \).

The power of complex numbers is in providing solutions to equations that have no real solutions, like \( x^2 + 1 = 0 \). This equation is solved when \( x = i \) and \( x = -i \). Therefore, understanding complex numbers is essential for solving a wide variety of mathematical problems.

Quadratic Equation

Quadratic equations are polynomial equations of the second degree, typically written in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations often involves finding the roots using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
For a quadratic equation, \( a \), \( b \), and \( c \) are coefficients and the discriminant, \( b^2 - 4ac \), determines the nature of the roots:

Real and Distinct Roots: When the discriminant is positive.
Real and Equal Roots: When the discriminant is zero.
Complex Roots: When the discriminant is negative.

In the given problem, the quadratic equation \( w^2 + w + 1 = 0 \) is solved using the quadratic formula. The discriminant here is \(-3\), leading to complex solutions \( w = \frac{-1 \pm i\sqrt{3}}{2} \). These solutions are further linked to the concept of roots of unity.

Euler's Formula

Euler's formula is a fundamental formula in complex analysis. It states \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), which beautifully ties together exponentials, trigonometric functions, and complex numbers.
This formula is particularly useful when working with exponential functions involving complex numbers, as it provides a way to express complex exponentials in terms of sines and cosines.

Connection to Trigonometry: It connects circular functions to exponentials.
Simplification of Calculations: It simplifies complex multiplication and division.

In solving the original exercise, Euler's formula helps convert the roots \( \omega = e^{i\frac{2\pi}{3}} \) and \( \omega^2 = e^{i\frac{4\pi}{3}} \) into solutions for \( z \). This conversion is crucial for understanding the periodic nature of complex exponentials.

Roots of Unity

Roots of unity are specific complex numbers which represent the distinct solutions of the equation \( z^n = 1 \). These roots are evenly spaced around the unit circle in the complex plane.
For example, the nth roots of unity are given by:
\[z_k = e^{\frac{2\pi i k}{n}},\quad \text{for } k = 0, 1, ..., n-1\]

Cube Roots of Unity: Involve \( n = 3 \), giving solutions for \( 1 \), \( \omega = e^{i\frac{2\pi}{3}} \), and \( \omega^2 = e^{i\frac{4\pi}{3}} \).
Geometric Interpretation: They form vertices of a regular polygon centered at the origin, inscribed in the unit circle.

In the exercise, \( \omega \) and \( \omega^2 \) were found to be solutions which indicate they are roots of unity, specifically the cube roots of unity. Understanding this concept is crucial, as it allows one to solve polynomial equations more easily, especially when linked to geometric properties.

91影视

In Problems \(33-36\), find all complex values of \(z\) satisfying the given equation. \(e^{2 z}+e^{z}+1=0\)

Short Answer

Step by step solution

Write the Equation in Terms of a Single Variable

Solve the Quadratic Equation

Find Values of \( z \) Using \( w = e^z \)

Key Concepts

Complex Numbers

Quadratic Equation

Euler's Formula

Roots of Unity

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