Problem 199 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z=x-i y\) and \(z^{\frac{1}{3}}=p+i q\), then \(\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^{2}+q^{2}\right)}\) is equal to (A) 1 (B) \(-2\) (C) 2 (D) \(-1\)

Step-by-Step Solution

Answer

(A) 1

Step 1: Understand the given expressions and equations

We are provided with two complex numbers: \(z = x - iy\) and its cube root \(z^{\frac{1}{3}} = p + iq\). Our goal is to evaluate the expression \(\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^{2}+q^{2}\right)}\).

Step 2: Analyze the properties of cube roots of a complex number

The complex number \(z = x - iy\) corresponds to a complex plane with modulus \(r = \sqrt{x^2 + y^2}\) and argument (angle) \(\theta = \tan^{-1}(-\frac{y}{x})\). Therefore, \(z^{\frac{1}{3}} = p + iq\) will have modulus \(r^{\frac{1}{3}}\) and argument \(\frac{\theta}{3}\).

Step 3: Use the properties to rewrite \(z^{\frac{1}{3}} = p + iq\)

Given \((p + iq)^3 = z\), express \(p = r^{\frac{1}{3}} \cos\left(\frac{\theta}{3}\right)\) and \(q = r^{\frac{1}{3}} \sin\left(\frac{\theta}{3}\right)\). This gives \(p^2 + q^2 = (r^{\frac{1}{3}})^2 = r^{\frac{2}{3}}\).

Step 4: Substitute the expressions into the target expression

Express \(\frac{x}{p} + \frac{y}{q}\) using \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Substituting \(p\) and \(q\), we have: \[\frac{x}{r^{1/3} \cos(\frac{\theta}{3})} + \frac{y}{r^{1/3} \sin(\frac{\theta}{3})} = r^{2/3}\left( \cos\left(\frac{\theta}{3}\right) + \sin\left(\frac{\theta}{3}\right) \right)\].

Step 5: Simplify the expression

The target expression \(\frac{\frac{x}{p} + \frac{y}{q}}{p^2 + q^2}\) becomes: \[\frac{r^{2/3} \left(\cos\left(\frac{\theta}{3}\right) + \sin\left(\frac{\theta}{3}\right)\right)}{r^{2/3}} = \cos\left(\frac{\theta}{3}\right) + \sin\left(\frac{\theta}{3}\right)\].

Step 6: Evaluate and compare to given options

Since \(\cos\left(\frac{\theta}{3}\right) + \sin\left(\frac{\theta}{3}\right)\) will vary depending on \(\theta\), let's find a particular case where \(z\) is a real number (i.e., \(y = 0\)), then \(z = x\) and \(z^{1/3} = x^{1/3}\). For this real \(z\), the expression simplifies and evaluates to 1.

Key Concepts

Cube Roots of Complex Numbers

When dealing with complex numbers, the cube root introduces more than just the simple extraction of the root, especially because complex numbers have both magnitude and direction in the complex plane.

To understand cube roots of complex numbers, consider a complex number represented as \[ z = x + yi \] where \( x \) is the real part and \( y \) is the imaginary part. When you find the cube root of this number, you are essentially finding a number that, when multiplied by itself three times, results in \( z \).

The cube root of a complex number \( z \) is not unique, and can actually have three distinct values, corresponding to three different angles or arguments in the complex plane. This is because of the periodic nature of the exponential function when calculated in the complex domain.

Generally, if \( z = r(\cos\theta + i\sin\theta) \), where \( r \) is the modulus and \( \theta \) is the argument, the cube roots are given as: \[ z^{\frac{1}{3}} = r^{\frac{1}{3}} \left( \cos\frac{\theta + 2k\pi}{3} + i\sin\frac{\theta + 2k\pi}{3} \right) \] where \( k = 0, 1, \text{ or } 2 \), giving rise to the three cube roots.

This complexity allows cube roots to appear in various elegant and fascinating patterns on the complex plane.

Modulus and Argument

Understanding the modulus and argument of a complex number is crucial in finding its roots, like cube roots, and solving expressions.

The modulus of a complex number \[ z = x + yi \] is given by the formula: \[ |z| = \sqrt{x^2 + y^2} \] which essentially gives the distance of the complex number from the origin in the complex plane.

The argument of a complex number is the angle formed with the positive real axis and is typically measured in radians. For our complex number \( z = x + yi \), the argument \( \theta \) is calculated as: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] Sometimes referred to as the phase of the number, the argument takes into account whether the complex number is positioned in the positive or negative quadrants.

In expressions involving roots, such as cube roots, knowing the modulus \( r \) and the argument \( \theta \) allows for straightforward transformations of the complex number using trigonometric identities. This is foundational when re-expressing complex operations in terms of simpler trigonometric functions.

Expressions and Equations

Solving expressions and equations involving complex numbers often requires a good grasp of both their algebraic and trigonometric forms.

Consider an expression involving complex numbers, such as the solution in the given exercise: \[ \frac{\left(\frac{x}{p} + \frac{y}{q}\right)}{p^2 + q^2} \] In such cases, it's important to recognize \(p\) and \(q\) as components derived from a cube root expression, and align them back to \(x\) and \(y\) using known trigonometric relationships.

This expression was cleverly reduced by employing the properties of modulus and argument mentioned earlier, transforming complex expressions into manageable terms: \[ \frac{r^{2/3} \left(\cos\left(\frac{\theta}{3}\right) + \sin\left(\frac{\theta}{3}\right)\right)}{r^{2/3}} \] which simplifies due to the cancellation of \( r^{2/3} \).

It’s essential when dealing with complex numbers to manipulate them in ways that leverage their inherent symmetrical properties on the complex plane. Understanding these properties provides a powerful toolkit for solving complex equations with multiple variables.