Problem 102 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(z_{1}, z_{2}\) are two complex numbers and \(w^{k}, k=0,1, \ldots\), \(n-1\) are the \(n\)th roots of unity, then \(\sum_{k=0}^{n-1}\left|z_{1}+z_{2} w^{k}\right|^{2}\) \((\mathrm{A})n\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)\) (D) can't say

Step-by-Step Solution

Answer

The answer is (B) \(= n(|z_1|^2 + |z_2|^2)\).

Step 1: Definition and Properties

To solve this problem, first recognize that the numbers \(w^k\) are the roots of unity, meaning \(w^n = 1\) and \(\sum_{k=0}^{n-1} w^k = 0\). These properties will be useful in simplifying the expression.

Step 2: Expand the Expression

We need to calculate \(\sum_{k=0}^{n-1} \left|z_1 + z_2 w^k\right|^2\). This can be expanded using the modulus squared property: \[ \left|z_1 + z_2 w^k\right|^2 = (z_1 + z_2 w^k)(\overline{z_1 + z_2 w^k}) = |z_1|^2 + z_1 \overline{z_2} \overline{w^k} + \overline{z_1} z_2 w^k + |z_2|^2 \].

Step 3: Summation over Roots of Unity

Sum the expanded expression over \(k\) from 0 to \(n-1\): \[ \sum_{k=0}^{n-1} \left[ |z_1|^2 + z_1 \overline{z_2} \overline{w^k} + \overline{z_1} z_2 w^k + |z_2|^2 \right] \].Using \(\sum_{k=0}^{n-1} w^k = 0\), the terms involving \(w^k\) simplify to zero:\[ \sum_{k=0}^{n-1} \left[ |z_1|^2 + |z_2|^2 \right] + 0 + 0 \] = \(n\left( |z_1|^2 + |z_2|^2 \right)\).

Step 4: Final Conclusion

Since the sum of the expression equals \(n\left( |z_1|^2 + |z_2|^2 \right)\), the correct answer is (B). Thus, the value of the sum is exactly \(n\left( |z_1|^2 + |z_2|^2 \right)\).

Key Concepts

Roots of Unity

In the complex number world, roots of unity hold significant importance. Imagine a unit circle within the complex plane. The roots of unity are points that lie on this circle, representing solutions to the equation \( w^n = 1 \). This equation means that \( w \) raised to the power of \( n \) equals 1. If you think of these roots as angles, they are evenly placed around the circle, each separated by \( \frac{2\pi}{n} \) radians.

A vital property associated with these roots is that their summation equals zero. This is because they are symmetrically distributed on the unit circle. For example, consider the third roots of unity, which are \( 1, \omega, \omega^2 \). Placing them in a summation, we get \( 1 + \omega + \omega^2 = 0 \). This zero sum plays a significant role in manipulating and simplifying expressions involving roots of unity, as seen in the provided problem's solution.

Modulus Squared

When dealing with complex numbers, often, you need to calculate their modulus. Think of modulus as the length or magnitude of a vector in a complex plane. The modulus squared, represented as \(|z|^2\), is particularly insightful. For a complex number \( z = a + bi \), its modulus squared is defined as \( a^2 + b^2 \).

Why is this concept useful? In expressions involving complex numbers, such as \( |z_1 + z_2 w^k|^2 \), understanding modulus squared helps simplify expressions analytically. Modulus squared converts the often challenging complex expressions into simpler arithmetic ones. When expanded, \( |z_1 + z_2 w^k|^2 = (z_1 + z_2 w^k) \times \overline{(z_1 + z_2 w^k)} \). This strategy is advantageous as it transforms complex numbers into a form that emphasizes their real-number components, making it easier to handle in calculations.

Summation of Complex Expressions

Summation is a very straightforward yet powerful mathematical operation, important even in complex numbers. When you encounter an expression like \( \sum_{k=0}^{n-1} \left|z_1 + z_2 w^k\right|^2 \), you are essentially adding up outcomes for different roots of unity.

In the given problem, after expanding the expression using modulus squared, summation comes into play. Each selected \( k \) in the root adds a term to the sum. With the summation over roots of unity, symmetry simplifies the calculations significantly. Parts of the expression containing \( w^k \) cancel out, thanks to their zero total sum property. Therefore, understanding these dual principles of symmetry and cancellation facilitates the simplification process, eventually leading us to the result \( n(|z_1|^2 + |z_2|^2) \), as found in the exercise solution. This step is critical for efficiently evaluating complex expressions across different scenarios in mathematics.