Problem 9 - A Complete Resource Book in Mathematics for JEE Main

Question

vIf \((\sqrt{3}+i)^{100}=2^{99}(a+i b)\), then \(b=\) (A) \(\sqrt{3}\) (B) \(\sqrt{2}\) (C) 1 (D) None of these

Step-by-Step Solution

Answer

The value of \( b \) is \( \sqrt{3} \) (Option A).

Step 1: Understand the Problem

We are asked to find the value of \( b \) in the equation \( (\sqrt{3}+i)^{100} = 2^{99}(a+i b) \). This involves expanding the left-hand complex expression and equating it to the given form.

Step 2: Rewrite Complex Number in Polar Form

Express the complex number \( \sqrt{3} + i \) in polar form. The modulus is \( \sqrt{ (\sqrt{3})^2 + 1^2 } = 2 \), and the argument (angle) \( \theta \) is \( \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \). Thus, \( \sqrt{3} + i = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \).

Step 3: Apply De Moivre's Theorem

Using De Moivre's theorem, compute \( (\sqrt{3} + i)^{100} = \left( 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right)^{100} = 2^{100} \left( \cos\left(\frac{100\pi}{6}\right) + i \sin\left(\frac{100\pi}{6}\right) \right) \).

Step 4: Simplify the Angle

Calculate the angle \( \frac{100\pi}{6} = \frac{50\pi}{3} \). Find this angle modulo \( 2\pi \) to simplify: \( \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3} \). Thus, \( \frac{50\pi}{3} \equiv \frac{2\pi}{3} \pmod{2\pi} \).

Step 5: Compute the Trigonometric Values

Use \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) to find the complex number. Substitute these values back to get \( 2^{100} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \).

Step 6: Equate and Solve for b

We have \( 2^{100} \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 2^{99}(a + i b) \). Dividing both sides by \( 2^{99} \), we get \( 2(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = a + i b \). Thus, \( b = \sqrt{3} \).

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a cornerstone in complex number theory. It provides a way to raise complex numbers in polar form to a power.
This is expressed as \((r(\cos \theta + i \sin \theta))^n = r^n(\cos(n \theta) + i \sin(n \theta))\). Here, \(r\) is the modulus, \(\theta\) is the argument, and \(n\) is the exponent.

• To use the theorem, start by converting the complex number to polar form.
• Apply De Moivre's Theorem to find the result of exponentiation.
• This method simplifies the multiplication process for large powers.

De Moivre's Theorem shows that complex exponentiation is related to rotating a vector in the complex plane.

Polar Form

Polar form is a way of expressing complex numbers. It highlights both the magnitude and direction of the complex number vector.
The polar form of a complex number \(a + bi\) is given by \(r(\cos \theta + i \sin \theta)\).

• \(r\) is the modulus, found using \(\sqrt{a^2 + b^2}\).
• \(\theta\) is the argument, calculated by \( \tan^{-1}(\frac{b}{a})\).

Polar form simplifies operations like multiplication, division, and taking powers of complex numbers. By using the angle, it visually represents the number's position in the complex plane, making it intuitive to understand complex figures.

Trigonometric Values

Trigonometric values are essential for converting and computing complex numbers in polar form. They allow us to transform the rectangular form \((a + bi)\) into \((r(\cos \theta + i \sin \theta))\).
Understanding key trigonometric values is important:

• The cosine function, \(\cos \theta\), gives the horizontal component.
• The sine function, \(\sin \theta\), gives the vertical component.

For instance, in our solution, we use \(\cos(\frac{2\pi}{3})\) and \(\sin(\frac{2\pi}{3})\) to find the complex number's real and imaginary parts in its final form.
Familiarity with trigonometric identities and unit circle values makes these calculations faster and more intuitive.

Complex Exponentiation

Complex exponentiation extends the process of raising numbers to powers but in the realm of complex numbers.
It is crucial in converting and simplifying large power calculations in complex numbers. When performing complex exponentiation, polar form and De Moivre's Theorem are powerful tools:

• Convert the complex number into polar form.
• Apply De Moivre's Theorem to handle the power operation efficiently.
• The factor \(r^n\) easily computes the resulting modulus, while \(n\theta\) simplifies the argument calculation.

By using these techniques, complex exponentiation calculations become much more manageable, revealing the diversity and elegance of complex number operations.