Problem 174 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The roots of the equation \(z^{4}+1=0\) are (A) \((\pm 1 \pm i)\) (B) \((\pm 2 \pm 2 i)\) (C) \(\frac{1}{\sqrt{2}}(\pm 1 \pm i)\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (C) \(\frac{1}{\sqrt{2}}(\pm 1 \pm i)\).

Step 1: Identify the form

The equation given is \(z^4 + 1 = 0\). Rewriting it gives \(z^4 = -1\). The number \(-1\) can be expressed in exponential form as \(e^{i\pi}\).

Step 2: Solve using De Moivre's Theorem

De Moivre's Theorem helps us find the roots of complex numbers. The equation can have solutions of the form \(z = e^{i(\pi/4 + k\pi/2)}\) where \(k = 0, 1, 2, 3\).

Step 3: Calculate individual roots

Substituting different values of \(k\) gives:- For \(k = 0\), \(z = e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\)- For \(k = 1\), \(z = e^{i3\pi/4} = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\)- For \(k = 2\), \(z = e^{i5\pi/4} = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\)- For \(k = 3\), \(z = e^{i7\pi/4} = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\)

Step 4: Match with options

From the roots \(\frac{1}{\sqrt{2}}(\pm 1 \pm i)\) obtained, match with given options. This matches with option (C).

Key Concepts

Roots of Equations

The roots of an equation are the solutions that satisfy it. When dealing with polynomials and equations involving complex numbers, such as the equation \(z^4 + 1 = 0\), discovering the roots requires a systematic approach. One standard method is transforming the equation. Here, \(z^4 = -1\), and expressing \(-1\) using exponential form can make the equation more manageable.

For complex equations, the roots are not always obvious integers or simple fractions. Instead, they often involve complex numbers, where both real and imaginary parts come into play. To solve for roots effectively, especially for powers other than 2 (like cubes or fourth powers), strategies such as leveraging formulas and theorems—such as De Moivre's Theorem—are essential.

Applying these methods allows us to find roots like \(\frac{1}{\sqrt{2}}(\pm 1 \pm i)\), showcasing how complex numbers provide a robust framework for solving seemingly intricate equations.

De Moivre's Theorem

De Moivre's Theorem is an invaluable tool in both trigonometry and complex number theory. It simplifies finding powers and roots of complex numbers expressed in polar form. The theorem is stated as: for a complex number \(z = r(\cos \theta + i\sin \theta)\), raising \(z\) to the power of \(n\) gives:

• \(z^n = r^n (\cos(n\theta) + i\sin(n\theta))\).

This theorem is instrumental when solving equations like \(z^4 + 1 = 0\), where straightforward algebraic methods fall short.

By representing the complex number \(z^4\) in polar form and solving \(z^4 = e^{i\pi}\), De Moivre's allows us to determine that \(z\) can be represented as \(e^{i(\pi/4 + k\pi/2)}\) for different integer values of \(k\). This demonstrates the rich interconnection between exponential functions and trigonometric expressions in complex analysis, ultimately leading us to compute the distinct roots precisely.

Polar Form of Complex Numbers

Complex numbers possess both a rectangular form (a + bi) and a polar form. The polar form provides a different perspective by expressing the number in terms of magnitude \(r\) and angle \(\theta\) (where \(r\) is the distance from the origin in the complex plane, and \(\theta\) is the angle from the positive x-axis). The polar form is:

• \(z = r(\cos \theta + i\sin \theta)\)

This change of perspective is valuable when engaging with operations like multiplication, division, and root extraction of complex numbers.

For the equation \(z^4 = -1\), expressing \(-1\) as \(e^{i\pi}\) aligns with the concept of converting complex numbers to polar form. With \(-1 = e^{i\pi}\), we explore roots by adjusting the angle \(\theta\) through various values indicating distinct rotations in the complex plane. These adjustments lead us to the four fourth roots, specifically detailing each polar representation through computational steps like those outlined using De Moivre's approach. The polar form not only guides us in solving the equation effectively but also enriches our conceptual understanding of complex number behavior across various transformations.