Chapter 17: Problem 38
In Problems 35-38, find all values of \(z\) satisfying the given equation. $$ e^{2 z}+e^{z}+1=0 $$
Short Answer
Expert verified
The solutions are \( z = i\frac{2\pi}{3} + 2\pi k \) and \( z = -i\frac{2\pi}{3} + 2\pi k \) for \( k \in \mathbb{Z} \).
Step by step solution
01
Rewriting the Equation
The given equation is \( e^{2z} + e^{z} + 1 = 0 \). Let's introduce a substitution to make it look simpler. Let \( u = e^{z} \), so the equation becomes \( u^2 + u + 1 = 0 \), which is a quadratic equation in \( u \).
02
Solving the Quadratic Equation
We have the quadratic equation \( u^2 + u + 1 = 0 \). The solutions to a quadratic equation \( au^2 + bu + c = 0 \) are given by the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \). Applying the formula, we find \( u = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \).
03
Finding Values of z
Since \( u = e^{z} \), solving \( e^{z} = \frac{-1 + i\sqrt{3}}{2} \) and \( e^{z} = \frac{-1 - i\sqrt{3}}{2} \), we need to find values of \( z \). These are complex numbers where \( e^{z} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) or \( e^{z} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \), both of which correspond to the complex number \( e^{i\frac{2\pi}{3}} \) and \( e^{-i\frac{2\pi}{3}} \).
04
Determine the General Solution
From the equation \( e^{z} = e^{i\theta} \), we have \( z = i\theta + 2\pi k \) for \( k \in \mathbb{Z} \) (because of the periodicity of the exponential function on the complex plane, which is \( e^{z} = e^{z + 2\pi i} \)). So \( z = i\frac{2\pi}{3} + 2\pi k \) and \( z = -i\frac{2\pi}{3} + 2\pi k \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Equations
Exponential equations are equations where variables appear in exponents. They play a key role in mathematics and understanding them can help solve various types of problems.
In the given problem, the equation is \( e^{2z} + e^{z} + 1 = 0 \). The essence of solving exponential equations often revolves around simplifying the expression by using substitutions.
In the given problem, the equation is \( e^{2z} + e^{z} + 1 = 0 \). The essence of solving exponential equations often revolves around simplifying the expression by using substitutions.
- By substituting \( u = e^{z} \), the exponential equation transforms to a polynomial equation.
- This substitution simplifies handling complex exponents.
Quadratic Equations
In mathematics, quadratic equations are fundamental and appear frequently, including in this exercise where the problem reduces to a quadratic form after substitution. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \).
The solution to any quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here:
This result indicates the solutions involve complex numbers, requiring further insight into complex number properties.
The solution to any quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here:
- \( a = 1 \)
- \( b = 1 \)
- \( c = 1 \)
This result indicates the solutions involve complex numbers, requiring further insight into complex number properties.
Complex Roots
Complex roots occur when solving quadratic equations with a negative discriminant, which is the expression under the square root in the quadratic formula \( b^2 - 4ac \). In this case, the discriminant is negative, leading to non-real solutions.
The complex numbers are crucial in this solution because:
By continuing from polar forms of complex roots, one finds \( z \) in terms of an imaginary component plus integer multiples of full rotations, illustrating the infinitely periodic nature of these solutions.
The complex numbers are crucial in this solution because:
- The roots \( \frac{-1 \pm i\sqrt{3}}{2} \) stem from the nature of complex numbers, which capture solutions that real numbers cannot.
- Each complex number can be expressed in polar form, helpful for solving exponential equations.
By continuing from polar forms of complex roots, one finds \( z \) in terms of an imaginary component plus integer multiples of full rotations, illustrating the infinitely periodic nature of these solutions.