Problem 38 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

If \(x\) is a positive integer, then \((1+i \sqrt{3})^{n}\) \(+(1-i \sqrt{3}) \mathrm{n}\) is equal to (a) \(2^{n-1} \cos \frac{n \pi}{3}\) (a) \(2^{n} \cos \frac{n \pi}{3}\) (c) \(2^{n+1} \cos \frac{n \pi}{3}\) (d) none of these

Step-by-Step Solution

Answer

The answer is (c) \(2^{n+1} \cos \frac{n\pi}{3}\).

Step 1: Express in Polar Form

Recognize that the complex numbers \(1 + i\sqrt{3}\) and \(1 - i\sqrt{3}\) can be expressed in polar form. Calculate their magnitudes to be 2. Therefore, \(1 + i\sqrt{3} = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\) and \(1 - i\sqrt{3} = 2\left(\cos\frac{-\pi}{3} + i\sin\frac{-\pi}{3}\right)\).

Step 2: Use De Moivre’s Theorem

Applying De Moivre’s Theorem to \((1+i\sqrt{3})^{n}\) and \((1-i\sqrt{3})^{n}\), we have: \((1+i\sqrt{3})^{n} = 2^n\left(\cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}\right)\) \((1-i\sqrt{3})^{n} = 2^n\left(\cos \frac{-n\pi}{3} + i \sin \frac{-n\pi}{3}\right)\).

Step 3: Add the Results

Add \((1 + i\sqrt{3})^{n}\) and \((1 - i\sqrt{3})^{n}\): \[ (1+i\sqrt{3})^{n} + (1-i\sqrt{3})^{n} = 2^n \left(\cos \frac{n\pi}{3} + i\sin \frac{n\pi}{3}\right) + 2^n \left(\cos \frac{-n\pi}{3} + i\sin \frac{-n\pi}{3}\right). \]

Step 4: Simplify the Expression

Notice that \(\sin\frac{n\pi}{3} + \sin\frac{-n\pi}{3} = 0\), so the imaginary parts cancel out. Thus, the sum simplifies to: \[ 2^n \left(\cos \frac{n\pi}{3} + \cos \frac{-n\pi}{3}\right) = 2^{n+1} \cos \frac{n\pi}{3}. \]

Step 5: Compare with Given Options

The resulting expression \(2^{n+1} \cos \frac{n\pi}{3}\) matches option (c). Thus, the answer to the problem is (c).

Key Concepts

Polar Form

Complex numbers are intriguing because they involve combining real and imaginary parts. When we express these numbers in rectangular form as \(a + bi\), where \(a\) and \(b\) are real numbers, it might sometimes be tricky to work with them directly in operations such as multiplication or exponentiation. Here's where polar form steps in.
To translate a complex number into its polar form, we use its magnitude (or modulus) and angle (also known as the argument). The magnitude \(r\) is calculated as \(\sqrt{a^2 + b^2}\), while the angle \(\theta\) stems from the tan inverse of \(b/a\). When expressed in polar form, a complex number looks like this:

• \(z = r (\cos \theta + i \sin \theta)\)

In our exercise, both complex numbers \(1 + i\sqrt{3}\) and \(1 - i\sqrt{3}\) are turned into polar form. With their magnitude calculated as \(2\), it simplifies to:

• \(1 + i\sqrt{3} = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)
• \(1 - i\sqrt{3} = 2\left(\cos\frac{-\pi}{3} + i\sin\frac{-\pi}{3}\right)\)

Using the polar form simplifies a lot of calculations, especially exponentiation, simplifying our working methods with complex numbers.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in mathematics, especially when it comes to working with complex numbers in polar form. It provides an efficient way to compute powers of complex numbers without having to multiply them out manually each time.
According to De Moivre's Theorem, for a complex number expressed in polar form \(r(\cos \theta + i \sin \theta)\), its \(n^{th}\) power can be found by raising the modulus \(r\) to the power \(n\) and multiplying the angle \(\theta\) by \(n\). Mathematically, it means:

• \((r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))\)

For our complex numbers \((1+i\sqrt{3})\) and \((1-i\sqrt{3})\), applying De Moivre's Theorem gives:

• \((1+i\sqrt{3})^n = 2^n(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3})\)
• \((1-i\sqrt{3})^n = 2^n(\cos\frac{-n\pi}{3} + i\sin\frac{-n\pi}{3})\)

This theorem simplifies our computations significantly, by avoiding repetitive multiplication and directly giving us a result in polar form based on known mathematical properties.

Integer Exponentiation

Exponentiation is a basic operation that involves raising a number to a power. In the context of complex numbers, exponentiation becomes interesting, especially when using their polar forms along with De Moivre's Theorem.
In the current exercise, we apply integer exponentiation to complex numbers expressed in polar form. The smooth integration of De Moivre’s Theorem simplifies the complex exponentiation. It involves raising both the radius to the integer power and multiplying the angle by the power.
Using De Moivre's Theorem in this exercise allows the transformation of intricate multiplication of complex numbers into manageable steps. For example, by raising \(2\) (the modulus of \(1 + i\sqrt{3}\)) to the power of \(n\), and multiplying \(\pi/3\) (the argument) by \(n\), excellent simplification is achieved:

• \((1+i\sqrt{3})^n = 2^n(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3})\)
• \((1-i\sqrt{3})^n = 2^n(\cos\frac{-n\pi}{3} + i\sin\frac{-n\pi}{3})\)

When we add these results, the imaginary parts, due to symmetry, cancel each other out. The cosine terms double up due to trigonometric identity, leading to the desired neat solution of\(2^{n+1} \cos \frac{n\pi}{3}\), exactly as required in option (c). This clear mathematical progression exemplifies the power and usefulness of integer exponentiation in solving complex numerical problems efficiently.