Problem 38 - Advanced Engineering Mathematics

Question

In Problems 35-38, find all values of \(z\) satisfying the given equation. $$ e^{2 z}+e^{z}+1=0 $$

Step-by-Step Solution

Answer

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 1: 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 \( u \).

Step 2: 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 \( u = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \).

Step 3: 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 \( e^{-i\frac{2\pi}{3}} \).

Step 4: 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 \( z = -i\frac{2\pi}{3} + 2\pi k \).

Key Concepts

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.

• By substituting \( u = e^{z} \), the exponential equation transforms to a polynomial equation.
• This substitution simplifies handling complex exponents.

This approach paves the way to apply different algebraic techniques, like solving quadratic equations. This simplification highlights how a seemingly complex exponential problem can be tackled by converting it into a more manageable form. In particular, it helps focus on the properties of exponential functions, including their behavior in the complex plane and periodicity, making further analysis straightforward.

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:

• \( a = 1 \)
• \( b = 1 \)
• \( c = 1 \)

After applying the formula to \( u^2 + u + 1 = 0 \), \( u \) is found to be complex, specifically \( u = \frac{-1 \pm i\sqrt{3}}{2} \).
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:

• 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.

Using Euler’s formula, \( e^{i\theta} = \cos \theta + i\sin \theta \), both roots provide ways to express potential angles. These angles, \( \frac{2\pi}{3} \) and \( -\frac{2\pi}{3} \), convert easily into expressions for \( e^{z} \), connecting back to the exponential form
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.