Problem 5 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

\(\left(\frac{\sqrt{3}+i}{2}\right)^{69}\) is equal to (a) 1 (b) \(-1\) (c) \(-i\) (d) \(i\)

Step-by-Step Solution

Answer

The expression \( \left( \frac{\sqrt{3}+i}{2} \right)^{69} \) equals \( i \).

Step 1: Evaluate the complex form

Begin by analyzing the expression \( \left( \frac{\sqrt{3}+i}{2} \right)^{69} \). Recognize that the expression represents a complex number in the form of a multiple angle of a unit circle. Identify \( \frac{\sqrt{3}}{2} + \frac{1}{2} i \) as a point on the unit circle, which can be connected to an angle using trigonometry.

Step 2: Convert to polar form

Convert the given complex number to polar form. The complex number \( \frac{\sqrt{3}}{2} + \frac{1}{2} i \) corresponds to the angle \( \theta = \frac{\pi}{6} \) because its cosine is \( \frac{\sqrt{3}}{2} \) and sine is \( \frac{1}{2} \). So it can be expressed in polar form as \( e^{i \frac{\pi}{6}} \).

Step 3: Apply De Moivre's Theorem

To find \( \left( \frac{\sqrt{3}+i}{2} \right)^{69} \), substitute the polar form into De Moivre's Theorem: \( \left( e^{i \frac{\pi}{6}} \right)^{69} = e^{i \cdot 69 \cdot \frac{\pi}{6}} \). This simplifies to \( e^{i \cdot \frac{69\pi}{6}} \).

Step 4: Simplify the angle

Reduce the angle \( \frac{69\pi}{6} \) modulo \( 2\pi \) to determine the corresponding angle on the unit circle. First, calculate \( \frac{69\pi}{6} = 11.5\pi \). Since \( 2\pi = 1 \) full circle, \( 11.5\pi \) is equivalent to \( 0.5 \pi \) beyond \( 5.5 \times 2\pi \). Therefore, the position is \( e^{i \cdot 0.5\pi} \).

Step 5: Convert simplified angle to complex form

The angle \( 0.5\pi \) corresponds to the point on the unit circle where cosine of \( 90^\circ \) or \( 0.5\pi \) is \( 0 \) and sine is \( 1 \), which translates to the complex number \( i \). Hence, \( e^{i \cdot 0.5\pi} = i \).

Key Concepts

Polar Form

Complex numbers can have both a rectangular form, like the numbers we’re used to working with, and a polar form, which can simplify multiplication and division of complex numbers.
The polar form of a complex number expresses it using an angle and a magnitude, often utilizing Euler’s formula.The formula to convert a complex number from its rectangular form, say \(a + bi\), to its polar form is given by:

• Magnitude: \( r = \sqrt{a^2 + b^2} \)
• Angle (Theta): \( \theta = \tan^{-1}(\frac{b}{a}) \)

In our example, the complex number \( \frac{\sqrt{3}}{2} + \frac{1}{2}i \) is expressed as a point on the unit circle. Its polar form uses an angle θ of \( \frac{\pi}{6} \), marking it on the unit circle as \( e^{i\frac{\pi}{6}} \).Converting complex numbers to polar form makes it easier to raise numbers to powers or to multiply them, which leads us to our next core concept.

De Moivre's Theorem

Once a complex number is in polar form, De Moivre's Theorem can be used to compute powers and roots of complex numbers efficiently.
This theorem states:\[(e^{i\theta})^n = e^{in\theta}\]This transformation means we can multiply the angle by n when raising to a power, which significantly simplifies the calculation.In the problem discussed, when the expression \( \left( e^{i \frac{\pi}{6}} \right)^{69} \) is considered, De Moivre’s Theorem helps convert it into \( e^{i \cdot 69 \cdot \frac{\pi}{6}} \).Thus, the task of raising complex numbers to even very large powers becomes manageable, as the multiplication of angles is far simpler than raising polynomials.

Unit Circle

The unit circle is a crucial concept in trigonometry and complex numbers, denoting a circle of radius one centered at the origin of the complex plane.
Each point on this circle corresponds to a complex number whose magnitude is one. It helps in understanding the cyclic nature of trigonometric functions and is integral in deriving angles for complex numbers.Points on the unit circle are defined by angles, where specific angles correspond to well-known complex numbers, like:

• \( \theta = 0 \) corresponds to the point (1, 0)
• \( \theta = \frac{\pi}{2} \) corresponds to the point (0, 1)
• \( \theta = \pi \) corresponds to the point (-1, 0)
• \( \theta = \frac{3\pi}{2} \) corresponds to the point (0, -1)

For the exercise, \( \frac{\pi}{6} \) places the complex number on the unit circle, simplifying its description to \( e^{i\frac{\pi}{6}} \) which aligns nicely with other calculations using polar coordinates and De Moivre's Theorem.