Problem 20 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solutions are \( z = -\frac{\pi}{4} + \pi n \), \( n \\in \\mathbb{Z} \).

Step 1: Understand the given equation

We have the equation \( \cos z = i \sin z \). Our goal is to find the values of \( z \) that satisfy this equation. We recognize that this involves trigonometric functions and their complex properties.

Step 2: Divide both sides by \( \cos z\)

Divide both sides of the equation by \( \cos z \) to isolate \( \tan z \): \[ 1 = i \tan z \] Now we have the equation \( \tan z = \frac{1}{i} \).

Step 3: Simplify \( \frac{1}{i} \)

Simplifying \( \frac{1}{i} \):\[ \frac{1}{i} = -i \] This means we have \( \tan z = -i \).

Step 4: Express \( \tan z \) using its exponential form

Recall that \( z = x + yi \), where \( x \) and \( y \) are real numbers. The complex tangent function is \[ \tan z = \frac{\sin z}{\cos z} = \frac{\frac{e^{iz} - e^{-iz}}{2i}}{\frac{e^{iz} + e^{-iz}}{2}} = -i \].

Step 5: Solve using the exponential form of \( \tan z \)

Use the exponential form for \( \tan z = -i \):Set \[ \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = -i \] Let's clear the \( i \) term:\[ i(e^{iz} - e^{-iz}) = -(e^{iz} + e^{-iz}) \] \[ ie^{iz} - ie^{-iz} = -e^{iz} - e^{-iz} \].

Step 6: Rearrange terms

Rewrite the equation: \[ ie^{iz} + e^{iz} = ie^{-iz} - e^{-iz} \] This implies \[ (i+1)e^{iz} = (i-1)e^{-iz} \].

Step 7: Find the solution for \( e^{iz} \)

Divide both terms by \((i-1)\):\[ e^{2iz} = \frac{i-1}{i+1} \].

Step 8: Simplify \( \frac{i-1}{i+1} \) using complex conjugates

Multiply the fraction by its conjugate:\[\frac{i-1}{i+1} \cdot \frac{i-1}{i-1} = \frac{(i-1)^2}{(i+1)(i-1)} = \frac{-2i}{2} = -i\] This gives \[ e^{2iz} = -i \].

Step 9: Solve for \( z \) using logarithms

Consider \[ e^{2iz} = e^{-\frac{\pi}{2}i + 2\pi ni} \] (since \(-i = e^{-\frac{\pi}{2}i}\)).Thus, \[ 2iz = -\frac{\pi}{2}i + 2\pi ni \].So dividing by 2:\[ z = -\frac{\pi}{4} + \pi n \].

Step 10: Conclude the solution

The values of \( z \) satisfying the equation \( \cos z = i \sin z \) are \[ z = -\frac{\pi}{4} + \pi n \]where \( n \in \mathbb{Z} \), the set of all integers.

Key Concepts

Trigonometric Equations

Trigonometric equations involve expressions with trigonometric functions such as sine, cosine, and tangent. These equations often require algebraic manipulations or substitutions to solve. Trigonometric identities are frequently employed to simplify these expressions.
In solving the equation \( \cos z = i \sin z \), we started by dividing both sides by \( \cos z \) to isolate \( \tan z \). This manipulation is typical in trigonometric problems to reduce the complexity by using known identities like \( \tan z = \frac{\sin z}{\cos z} \).
Once simplified, solving trigonometric equations with complex numbers frequently involves translating the expression into an exponential form for easier handling.

Complex Analysis

Complex analysis is a powerful field of mathematics focused on complex numbers and functions involving them. It's particularly useful in understanding phenomena that cannot be captured adequately with real numbers alone.
In our task of solving \( \cos z = i \sin z \), complex analysis allows us to simplify using exponential forms. By expressing \( \tan z \) and later the equation \( e^{iz} \), we utilize the principles of complex function theory. This approach shows the inherent beauty in complex algebra, where growth and oscillations can be analyzed cleanly through exponential expressions.

Exponential Form of Complex Numbers

The exponential form of complex numbers provides a compact representation, making calculations involving trigonometric and hyperbolic functions more manageable. A complex number can be expressed as \( e^{i\theta} = \cos \theta + i \sin \theta \), which beautifully connects trigonometric functions with exponential growth.
In our solution, we used exponential forms to handle \( e^{iz} \). We converted complex trigonometric expressions into a more workable form by leveraging Euler's formula. This simplification is vital for solving equations like \( \cos z = i \sin z \), ultimately leading to cleaner and more traceable results.

Imaginary Unit

The imaginary unit, denoted as \( i \), is essential in complex number arithmetic. It represents the square root of \( -1 \), enabling extensive calculations that transcend the limitations of real numbers. The imaginary unit allows us to capture rotations and oscillations, described naturally by sine and cosine functions.
In the given problem, the appearance of \( i \sin z \) indicates rotation by \( 90^\circ \), typical in equations describing waveforms or rotations in complex analysis. Understanding \( i \) as multiplying by \(-i\) mimics a rotation in the opposite direction, which was crucial in simplifying \( \frac{1}{i} \) as \(-i\). Mastering these interpretations is essential for accommodating rotational dynamics in trigonometric equations.