Problem 107 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(i=\sqrt{-1}\), then \(4+5\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{365}\) is equal to (A) \(1-i \sqrt{3}\) (B) \(-1+i \sqrt{3}\) (C) \(i \sqrt{3}\) (D) \(-\sqrt{3} i\)

Step-by-Step Solution

Answer

The expression simplifies to (D) \(-\sqrt{3} i\).

Step 1: Identify the Complex Root

The complex number \[a = -\frac{1}{2} + \frac{i \sqrt{3}}{2}\]is a complex root of unity. Specifically, it is one of the cube roots of unity, as \[a^3 = 1.\]We can use the properties of cube roots of unity to simplify further powers of \(a\).

Step 2: Simplifying Powers

Given that \[a^3 = 1,\]we can express higher powers of \(a\) using the relation \[a^n = a^{n \bmod 3}.\]Thus, we compute:\[334 \bmod 3 = 2, \quad \text{so} \quad a^{334} = a^2.\]\[365 \bmod 3 = 2, \quad \text{so} \quad a^{365} = a^2.\]Both simplify to \(a^2\).

Step 3: Calculate \(a^2\)

Calculate \(a^2\) using:\[a = -\frac{1}{2} + \frac{i \sqrt{3}}{2}.\]Squaring it yields:\[a^2 = \left(-\frac{1}{2} + \frac{i \sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{\sqrt{3}i}{2} - 3 \cdot \frac{1}{4} = -\frac{1}{2} - \frac{i \sqrt{3}}{2}.\]

Step 4: Substitute and Simplify the Expression

Substitute \( a^2 = -\frac{1}{2} - \frac{i \sqrt{3}}{2} \) into the given expression:\[4 + 5a^{334} + 3a^{365} = 4 + 5 \left( -\frac{1}{2} - \frac{i \sqrt{3}}{2} \right) + 3\left(-\frac{1}{2} - \frac{i \sqrt{3}}{2}\right).\]Simplify:\[5 \times -\frac{1}{2} + 3 \times -\frac{1}{2} = -4\]and\[\5 \left(-\frac{i \sqrt{3}}{2}\right) + 3 \left(-\frac{i \sqrt{3}}{2}\right) = -4i \sqrt{3}\]Combine the real and imaginary parts:\[\4 - 4 = 0\]\[0 - 4i \sqrt{3} = -4i\sqrt{3}.\]

Step 5: Concluding the Result

The simplified expression evaluates to \[-4i \sqrt{3},\]which matches option (D) \[-\sqrt{3} i.\]The expression simplifies to the choice given by \(-\sqrt{3} i\).

Key Concepts

Cube Roots of Unity

Cube roots of unity are fascinating elements in the world of complex numbers. They are the three solutions to the equation \[ x^3 = 1 \] In simpler terms, these roots are the numbers that, when raised to the third power, equal one. For cube roots of unity, you will always have:

• 1: The trivial root.
• \( \omega = -\frac{1}{2} + \frac{i\sqrt{3}}{2} \) which is a nontrivial root.
• \( \omega^2 = -\frac{1}{2} - \frac{i\sqrt{3}}{2} \) which is the square of the nontrivial root.

Note that \( \omega^3 = 1 \), making it a cube root of unity. These roots are evenly spaced on the complex plane, forming a perfect triangle surrounding the origin. They are useful in simplifying powers and manipulating expressions where rotation in the complex plane is significant.
In the exercise, recognizing the complex number as a cube root of unity helps to reduce the complexity of higher powers. This simplifies expressions significantly when dealing with polynomial equations or when continuous rotations or periodicities are involved.

Powers of Complex Numbers

Complex numbers can be daunting, yet they offer a beautiful expanse of mathematical functionality, especially when considering their powers. Any complex number can be expressed in the form \[ z = a + bi \]where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit.
When analyzing powers of these complex numbers, particularly those involving cube roots of unity like \( \omega = -\frac{1}{2} + \frac{i\sqrt{3}}{2} \),understanding that their powers rotate in a cycle is key. Despite seeming complex, these numbers follow a simple pattern due to their geometric nature on the complex plane.
With cube roots of unity, the powers cycle every three terms. This means,

• \( \omega^1 = \omega \)
• \( \omega^2 \) becomes \( \omega^2 \)
• \( \omega^3 = 1 \)
• \( \omega^4 = \omega \)

This cyclical nature allows for easy computation by reducing large exponents to manageable numbers using modular arithmetic, which simplifies expressions tremendously in calculations.

Modular Arithmetic

Modular arithmetic, often considered a refined version of 'math clock,' simplifies calculations on a cyclic scale. It deals with remainders and is indispensable in solving problems involving powers of numbers, especially in the realm of complex numbers.
For instance, when calculating powers of complex numbers like the cube roots of unity, knowing \[ a^3 = 1 \]lets you use modular arithmetic to simplify the exponent. Instead of calculating enormous powers, such as \( a^{334} \), use:

• \( 334 \mod 3 = 2 \)
• \( 365 \mod 3 = 2 \)

Thus, both \( a^{334} \) and \( a^{365} \) reduce to \( a^2 \). This brings a huge simplification and allows you to avoid lengthy direct computation, offering a more efficient pathway to reach the correct answer without missing important steps.

Imaginary Unit

Imaginary units let mathematicians extend the concept of numbers beyond real values to solve equations that have no real solutions. The imaginary unit, denoted as \( i \), is defined such that:\[ i = \sqrt{-1} \]This peculiar property allows algebra to address square roots of negative numbers and expand into complex numbers, forming the base of complex arithmetic.
In complex numbers, \( i \) is crucial as a component of numbers like \( a = -\frac{1}{2} + \frac{i \sqrt{3}}{2} \). Such numbers have both a real and an imaginary part, enabling operations to go beyond the limitations of real numbers alone.

• \( i^2 = -1 \)
• \( i^3 = -i \)
• \( i^4 = 1 \)

These identities repeat in a cycle, proving useful in simplifying powers of \( i \), much like what happens with cube roots of unity. The imaginary unit allows exploration of polynomial solutions, electrical engineering, wave functions, and more, rendering mathematics more comprehensive and versatile.