Problem 16 - Advanced Engineering Mathematics

Question

Find all values of \(z\) satisfying the given equation. \(\cos z=-3 i\)

Step-by-Step Solution

Answer

Given the complex nature of cosine, solutions for \(z\) involve using the general formula \(z = -i \ln(w) + 2k\pi\) for each potential \(w\), where \(k\) is any integer.

Step 1: Express General Solution for Cosine

For a complex number in exponential form, the cosine can be found using the formula: \( \cos z = \frac{e^{iz} + e^{-iz}}{2} \). Given that \( \cos z = -3i \), we set up the equation: \[ \frac{e^{iz} + e^{-iz}}{2} = -3i \].

Step 2: Clear Fractions

Multiply the entire equation by 2 to remove the fraction: \[ e^{iz} + e^{-iz} = -6i \].

Step 3: Introduce Substitution

Let \( w = e^{iz} \). Then \( e^{-iz} = \frac{1}{w} \), and the equation becomes \[ w + \frac{1}{w} = -6i \].

Step 4: Multiply by \(w\)

To eliminate the fraction, multiply the entire equation by \( w \): \[ w^2 + 1 = -6iw \].

Step 5: Rearrange to Quadratic Form

Rearrange to form a standard quadratic equation: \[ w^2 + 6iw + 1 = 0 \].

Step 6: Solve Quadratic Equation

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = 6i, c = 1 \). This gives: \[ w = \frac{-6i \pm \sqrt{(6i)^2 - 4}}{2} \]. Calculate \( (6i)^2 - 4 = -36 - 4 = -40 \), thus: \[ w = \frac{-6i \pm \sqrt{-40}}{2} \].

Step 7: Simplify the Radical Term

Simplify \( \sqrt{-40} = \sqrt{40} \cdot i = 2 \sqrt{10} i \). The solutions for \( w \) are: \[ w = \frac{-6i \pm 2 \sqrt{10} i}{2} \], which simplifies to \[ w = -3i \pm \sqrt{10} i \].

Step 8: Find \(z\) for Each \(w\)

We have two possible values for \(w\): \(-3 - \sqrt{10} \) and \(-3 + \sqrt{10} \). Since \( w = e^{iz} = \cos(\theta) + i \sin(\theta) \), calculate \( z \) by using the inverse cosine method for each solution.

Step 9: Express General Solution

Since \( w = e^{iz} \), correspondingly \( z = -i \ln(w) \). Calculate the logarithm for each potential \(w\) considering the periodicity of the complex exponential function.

Key Concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are often written in the form: \[ a + bi \] where \( a \) is the real part and \( b \) is the imaginary part multiplied by \( i \), which satisfies \( i^2 = -1 \). This representation allows us to work with numbers in a two-dimensional plane called the complex plane.

• The horizontal axis represents the real part of the number.
• The vertical axis represents the imaginary part.

Complex numbers enable the expression of mathematical phenomena that cannot be explained using only real numbers. They are fundamental to various fields of mathematics and engineering. When dealing with equations involving complex numbers, we can often utilize exponential forms to simplify solutions, as the relationship between exponential forms and trigonometric functions is a key aspect of complex analysis.

Quadratic Equations

Quadratic equations play a significant role in complex algebra. A typical quadratic equation has the form: \[ ax^2 + bx + c = 0 \] where \( a, b, \) and \( c \) are constants, and \( x \) represents an unknown variable. The solutions to a quadratic equation can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula computes the roots or solutions of the quadratic equation, which could be real or complex.
In the context of complex numbers, quadratic equations can yield complex solutions, as demonstrated in the provided problem. The roots are evaluated by performing calculations that involve both real and imaginary parts, often requiring simplification of terms like \( \sqrt{-1} \) into forms where they can be easily handled. Determining the roots accurately is crucial since these values serve as the basis for further computations, such as finding inverse functions.

Exponential Form of Complex Numbers

The exponential form of complex numbers is a more compact way of expressing complex numbers, especially useful in simplifying trigonometric calculations. Euler's formula is foundational in this, stating: \[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \] This representation links exponential functions with trigonometric functions seamlessly.
The form leverages the exponential function \( e^{iz} \) to represent a complex number \( z \) as a point in the complex plane. The exponential form is not only elegant but also enables rotation and magnification of complex numbers in a straightforward manner by multiplying exponential terms. It transforms problems involving \( \cos z \) and \( \sin z \) into simpler calculations involving powers and products of complex numbers, as seen in the exercise, where solving for \( \cos z \) becomes a matter of handling exponential powers. Using this form helps find solutions to complex problems more efficiently, including finding multiple solutions due to the periodicity of trigonometric functions.

Inverse Functions

Inverse functions are operations that `undo' each other. For complex numbers, finding the inverse involves understanding operations such as logarithms and trigonometric inverses in the complex domain. When you have a function \( f(x) \), its inverse \( f^{-1}(x) \) satisfies: \[ f(f^{-1}(x)) = x \]
To solve for \( z \) from \( w = e^{iz} \), we take the natural logarithm of both sides, considering \( e^{iz} \) as the inverse of \( \ln(w) \) in the complex plane. While the real logarithm function is typically straightforward, complex logarithms involve a periodic component because of the properties of \( e^{iz} \), leading to multiple valid logarithms differing by multiples of \( 2\pi i \). This periodicity must be taken into account when computing inverse trigonometric functions, such as \( \cos^{-1}(x) \). Understanding these inverse relationships is vital when resolving equations involving complex exponentials, ensuring all potential solutions are identified, mirroring how the inverse sine and cosine functions work in real analysis.