Problem 22 - Advanced Engineering Mathematics

Question

\(\sin z=i \sinh 2\) implies \(\sin x \cosh y+i \cos x \sinh y=0+i \sinh 2\) and so we must have \(\sin x \cosh y=0\) and \(\cos x \sinh y=\sinh 2 .\) since \(\cosh y>0\) for all real numbers, the first equation has only the solutions \(x=n \pi\) \(n=0,\pm 1,\pm 2, \ldots .\) For \(x=n \pi\) the second equation becomes \((-1)^{n} \sinh y=\sinh 2 .\) If \(n\) is even, \(\sinh y=\sinh 2\) implies \(y=2(\sinh y \text { is one-to-one.) If } n \text { is odd, }-\sinh y=\sinh 2 \text { implies } \sinh y=-\sinh (-2) \text { and so } y=-2\) Solutions of the original equation are then $$z=2 k \pi+2 i, \quad z=(2 k+1) \pi-2 i, \quad k=0,\pm 1,\pm 2, \dots$$.

Step-by-Step Solution

Answer

Solutions are \(z = 2k\pi + 2i\) and \(z = (2k + 1)\pi - 2i\) for \(k\in \mathbb{Z}\).

Step 1: Interpret the Given Equation

We start with the equation \(\sin z = i \sinh 2\). This implicity defines a relationship between the real and imaginary components of a complex number \(z = x + iy\). By equating real and imaginary parts, the equation is split into two separate equations.

Step 2: Split into Real and Imaginary Equations

Given \(z = x + iy\), the expression for the sine of a complex number is \(\sin z = \sin x \cosh y + i \cos x \sinh y\). Thus, we equate this to \(0 + i \sinh 2\):1. \(\sin x \cosh y = 0\)2. \(\cos x \sinh y = \sinh 2\)

Step 3: Solve the First Equation

The equation \(\sin x \cosh y = 0\) implies either \(\sin x = 0\) or \(\cosh y = 0\). However, since \(\cosh y > 0\) for all real numbers, we have \(\sin x = 0\). Therefore, \(x = n\pi\) where \(n = 0, \pm 1, \pm 2, \ldots\).

Step 4: Solve the Second Equation for \(x = n\pi\)

Substituting \(x = n\pi\) into the second equation \(\cos x \sinh y = \sinh 2\), we get \((-1)^n \sinh y = \sinh 2\).

Step 5: Determine the Values of \(y\)

Depending on the parity of \(n\):- If \(n\) is even, \(\sinh y = \sinh 2\), hence \(y = 2\) since the hyperbolic sine function is one-to-one.- If \(n\) is odd, \(-\sinh y = \sinh 2\), which simplifies to \(\sinh y = \sinh (-2)\), therefore \(y = -2\).

Step 6: Write the Solutions for \(z\)

Combine the values of \(x\) and \(y\) to write the solutions for \(z\):- For even \(n = 2k\), \(z = 2k\pi + 2i\).- For odd \(n = 2k + 1\), \(z = (2k + 1)\pi - 2i\).Thus, the solutions are \(z = 2k\pi + 2i\) and \(z = (2k + 1)\pi - 2i\) for \(k = 0, \pm 1, \pm 2, \ldots\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions, similar to trigonometric functions, are indivisible in understanding complex numbers. They appear in mathematical contexts like calculus and differential equations. The primary hyperbolic functions are the hyperbolic sine (\(\sinh x\)) and the hyperbolic cosine (\(\cosh x\)).
Here's how they are defined:

• Hyperbolic sine: \(\sinh x = \frac{e^x - e^{-x}}{2}\)
• Hyperbolic cosine: \(\cosh x = \frac{e^x + e^{-x}}{2}\)

Hyperbolic functions are closely linked to exponential functions. They exhibit properties that are quite analogous to those of trigonometric functions. For instance, a fundamental identity is:\[ \cosh^2 x - \sinh^2 x = 1 \]This mirrors the Pythagorean identity for trigonometric functions: \(\cos^2 x + \sin^2 x = 1\). Just as \cosh xis always positive, \sinh yis a function that uniquely defines real values
when given real numbers. This uniqueness is crucial when solving equations involving hyperbolic functions, especially when determining the possibility of \(\sinh y = \sinh 2 \) having specific solutions like \(y = 2\).

Imaginary and Real Components

Complex numbers are expressed in the form \(z = x + iy\), where \(x\) and \(y\) are real numbers representing the real and imaginary parts. Analyzing the real and imaginary components separately is a pivotal strategy in solving complex equations.
Consider the function \(\sin z = \sin x \cosh y + i\cos x \sinh y\):

• The real part is \(\sin x \cosh y\)
• The imaginary part is \(\cos x \sinh y\)

When presented with an equation such as \(\sin z = i\sinh 2\), the first step is to equate the real and imaginary components to zero and \(\sinh 2\), respectively. This separation allows for solving easier equations independently, revealing specific values for \(x\) and \(y\).
For instance, since the real part must be zero, solving \(\sin x \cosh y = 0\) simplifies our task to determining when \(\sin x = 0\), thus finding that \(x = n\pi\). For the imaginary part, \(\cos x \sinh y = \sinh 2\) guides us to specific solutions based on the calculated parity of \(n\). This method lets us dissect complex equations into manageable pieces.

Trigonometric Equations

Trigonometric equations involve angles and are essential in various scientific and engineering applications. When working with complex numbers, trigonometric equations are often encountered in their adapted forms for the complex plane.
Understanding how to manipulate these equations is key. Consider \(\sin z = i\sinh 2\). By splitting \(z = x + iy\) into its real and imaginary trigonometric components, we come across familiar trigonometric forms like \(\sin x = 0\).

• When \(\sin x = 0\), \(x = n\pi\), signifying multiple angle solutions that are crucial in periodicity.

Trigonometric properties such as period and identity are leveraged here. The sine function's periodic nature dictates that solutions recur at intervals of \(\pi\). Furthermore, solving \(\cos x \sinh y = \sinh 2\), hinges on substituting \(x = n\pi\), neatly tying the solutions to their hyperbolic counterparts.
Trigonometric equations in the context of complex numbers often lead to infinite possible solutions, structured around these periodic instances. Therefore, recognizing patterns in trigonometric equations alongside their hyperbolic pairs is crucial in composing and finding solutions, like separable real and imaginary parts of \(z\).