Question

The complex numbers \(\sin x+i \cos 2 x\) and \(\cos x-i \sin 2 x\) are conjugate of each other for (a) \(\mathrm{x}=\mathrm{k} \pi, \mathrm{k} \in \mathrm{z}\) (b) \(x=0\) (c) \(\mathrm{x}=[\mathrm{k}+(1 / 2)] \pi, \mathrm{k} \in \mathrm{z}\) (d) no value of \(\mathrm{x}\)

Short answer

The correct answer is (d) no value of x, as there is no intersection between the sets of solutions for the given system of equations.

Step-by-step solution

Conjugate of b

The conjugate of a complex number \(c = p + iq\) (where p and q are real numbers) is defined as \(\bar{c} = p - iq\). Therefore, the conjugate of \(b = \cos x - i\sin 2x\) is \(\bar{b} = \cos x + i\sin 2x\).

Set a equal to the conjugate of b

We need to find for which values of x, \(a = \bar{b}\), where \(a = \sin x + i\cos 2x\) and \(\bar{b} = \cos x + i\sin 2x\). Thus, we need to solve the equation \(\sin x + i\cos 2x = \cos x + i\sin 2x\).

Compare real and imaginary parts

In order for the equation \(\sin x + i\cos 2x = \cos x + i\sin 2x\) to be true, the real and imaginary parts must be equal. Therefore, we get the following system of equations: 1. \(\sin x = \cos x\) 2. \(\cos 2x = \sin 2x\)

Solve the system of equations

We need to find the values of x that satisfy both equations. Let's solve each equation separately. 1. \(\sin x = \cos x\). Since \(\sin x = \cos x\), we have \(\tan x = 1\). We know that \(\tan x = 1\) for \(x = \frac{\pi}{4} + k\pi\), where k is any integer. So, the solution to the first equation is \(x = \frac{\pi}{4} + k\pi, k \in \mathbb{Z}\). 2. \(\cos 2x = \sin 2x\). Since \(\cos 2x = \sin 2x\), we have \(\tan 2x = 1\). We know that \(\tan x = 1\) for \(x = \frac{\pi}{4} + k\pi\), where k is any integer. Thus, the solution to the second equation is \(x = \frac{\pi}{8} + \frac{k\pi}{2}, k \in \mathbb{Z}\).

Intersection of solutions

Now, we need to find the intersection of the solutions found in Step 4, which means the values of x that satisfy both equations. There is no intersection between these two sets of solutions. Therefore, there is no value of x for which \(\sin x + i\cos 2x\) and \(\cos x - i\sin 2x\) are conjugate of each other. The correct answer is (d) no value of x.

Key concepts

Complex Conjugate

Let's start understanding the complex conjugate concept. In mathematics, a complex number is of the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with \(i^2 = -1\). The complex conjugate of \(z\), denoted by \(\overline{z}\), is obtained by changing the sign of the imaginary part. Thus, the complex conjugate is given by \(\overline{z} = a - bi\).
A key property of complex conjugates is useful in complex arithmetic because multiplying a complex number by its conjugate results in a real number: \(z \cdot \overline{z} = (a + bi)(a - bi) = a^2 + b^2\).
This property makes them particularly useful when simplifying expressions involving complex numbers or when finding the modulus. In context, given \(b = \cos x - i \sin 2x\), its complex conjugate \(\overline{b} = \cos x + i \sin 2x\) is needed to solve the problem.

Trigonometric Identities

Trigonometric identities are essential tools in mathematics, especially when dealing with equations involving trigonometric functions like sine (\sin) and cosine (\cos).
These identities express relationships between different trigonometric functions, allowing us to simplify and solve equations.
Some key identities that are relevant in this problem include:

• \(\sin x = \cos x\) implies \(\tan x = 1\), which results from the identity \(\tan x = \frac{\sin x}{\cos x}\)
• \(\cos 2x = \sin 2x\) implies \(\tan 2x = 1\)

These identities help in linking the trigonometric equations to find solutions. Solving \(\sin x = \cos x\), for example, brings us to \(x = \frac{\pi}{4} + k\pi\), and solving \(\cos 2x = \sin 2x\) gives \(x = \frac{\pi}{8} + \frac{k\pi}{2}\).
Understanding trigonometric identities is vital for such problems as they allow simplification and solution of trigonometric equations.

Solution of Equations

Solving equations involves finding the values for which the equation holds true. In this exercise, we are given two conditions to find the values of \(x\) that make \(\sin x + i\cos 2x = \cos x + i\sin 2x\).
To solve such an equation, we equate both the real and imaginary parts separately. This gives a system of two equations:

• \(\sin x = \cos x\)
• \(\cos 2x = \sin 2x\)

Solving the first equation \(\sin x = \cos x\) results in \(x = \frac{\pi}{4} + k\pi\).
For the second, \(\cos 2x = \sin 2x\), it simplifies to \(\tan 2x = 1\), giving solutions \(x = \frac{\pi}{8} + \frac{k\pi}{2}\).
Each equation provides a series of solutions, known as a solution set. Understanding how to solve such trigonometric equations step-by-step requires a strong grasp of the identities and properties of these functions.

Intersection of Solutions

Finding the intersection of solutions means identifying where two or more sets of solutions overlap. In simpler terms, it is the common values that satisfy multiple equations simultaneously.
In our exercise, we need to determine when both \(\sin x + i\cos 2x = \cos x + i\sin 2x\) hold true using their isolated solutions:

• For \(\sin x = \cos x\), the solution set is \(x = \frac{\pi}{4} + k\pi\).
• For \(\cos 2x = \sin 2x\), the solution set is \(x = \frac{\pi}{8} + \frac{k\pi}{2}\).

We compare these sets to find a common x-value, if any.
Here, these two solution sets have no common values, meaning there is no x that satisfies both equations. Hence, there is no intersection between them.
This process highlights the importance of understanding how separate equations relate to finding where they intersect.