Problem 198 - A Complete Resource Book in Mathematics for JEE Main

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

(B) -2

Step 1: Understand the given information

We are given that if \(z = x - iy\) and \(z^{\frac{1}{3}} = p + iq\), we need to find the value of \( \frac{\left(\frac{x}{p} + \frac{y}{q}\right)}{\left(p^{2}+q^{2}\right)} \).

Step 2: Recognize the relationship between z and its cube root

Since \(z = (p + iq)^3\), we expand to get \(z = (p+iq)^3 = p^3 + 3p^2iq - 3pq^2 - iq^3 = (p^3 - 3pq^2) + i(3p^2q - q^3)\).

Step 3: Equate real and imaginary parts

From \(z = x - iy\) and \(z = (p^3 - 3pq^2) + i(3p^2q - q^3)\), we have the equations: \(x = p^3 - 3pq^2\) and \(-y = 3p^2q - q^3\).

Step 4: Substitute for x and y in the given expression

Substitute \(x = p^3 - 3pq^2\) and \(y = -(3p^2q - q^3)\) into the expression: \(\frac{\left(\frac{x}{p} + \frac{y}{q}\right)}{\left(p^2+q^2\right)} = \frac{\left(\frac{p^3 - 3pq^2}{p} + \frac{-(3p^2q - q^3)}{q}\right)}{\left(p^2+q^2\right)}\).

Step 5: Simplify the expression

Simplify inside the numerator: \(\frac{p^3 - 3pq^2}{p} = p^2 - 3q^2\) and \(\frac{-(3p^2q - q^3)}{q} = -3p^2 + q^2\). Add these: \( (p^2 - 3q^2) + (-3p^2 + q^2) = -2(p^2 + q^2)\).

Step 6: Final simplification

Now the expression becomes: \(\frac{-2(p^2 + q^2)}{p^2+q^2} = -2\).

Step 7: Conclusion

The expression simplifies to \(-2\), thus the answer is \(\text{B}\) which is \(-2\).

Key Concepts

Cube Roots

When dealing with complex numbers, the concept of cube roots can initially appear perplexing. Simply put, a cube root of a number is a value that, when raised to the power of three, returns the original number. For complex numbers, this involves not only determining the magnitude but also the direction or 'angle' of the number in the plane.

• In our given exercise, we have the complex number \(z = x - iy\) and its cube root \(z^{\frac{1}{3}} = p + iq\).
• We express the cube of \(p + iq\) to relate it back to the original \(z\).

Understanding this relationship is crucial because it allows us to work backwards, finding each component of \(z\) through its cube roots, and ultimately simplifying complex expressions.

Real and Imaginary Parts

Complex numbers have both real and imaginary parts, which are essential for their representation. The real part is a number without an imaginary unit, while the imaginary part includes the imaginary unit \(i\), representing \(\sqrt{-1}\).

• In our equations, \(z = x - iy\) and \(z = (p^3 - 3pq^2) + i(3p^2q - q^3)\), we equate the real parts \(x\) and \(p^3 - 3pq^2\), and the imaginary parts \(-y\) and \(3p^2q - q^3\).
• By separating these parts, we gain a clearer understanding of complex numbers, allowing us to solve the exercise systematically.

This separation is fundamental in complex arithmetic and ensures we know exactly how each part of the number contributes to its overall value.

Simplifying Expressions

Simplifying expressions involving complex numbers ensures they can be managed effectively in calculations. It involves reducing expressions to the simplest form where they are easier to work with, especially in problem-solving.

• In our solution, after substituting the values for \(x\) and \(y\), the goal was to simplify the overall expression \(\frac{(p^2 - 3q^2) + (-3p^2 + q^2)}{p^2+q^2}\).
• We recognized that \((p^2 - 3q^2) + (-3p^2 + q^2) = -2(p^2 + q^2)\), which simplifies to \(-2\) when the denominator \(p^2+q^2\) is the same.

This simplification lets us manage complex algebra seamlessly, breaking down seemingly intricate expressions into understandable results.