Problem 196 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\left(\frac{1+i}{1-i}\right)^{x}=1\), then (A) \(x=4 n\), where \(n\) is any positive integer (B) \(x=2 n\), where \(n\) is any positive integer (C) \(x=4 n+1\), where \(n\) is any positive integer (D) \(x=2 n+1\), where \(n\) is any positive integer

Step-by-Step Solution

Answer

(A) x = 4n, where n is any positive integer.

Step 1: Express the complex number in polar form

The given expression is \(\frac{1+i}{1-i}\). To simplify, multiply the numerator and denominator by the conjugate of the denominator: \(\frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{(1+2i+i^2)}{1-i^2}\). Simplify further: \(i^2 = -1\), so the expression becomes \(\frac{1+2i-1}{1+1} = \frac{2i}{2} = i\). Thus, \(\frac{1+i}{1-i} = i\).

Step 2: Solve for x using polar form

The polar form of \(i\) is \(e^{i\frac{\pi}{2}}\) since it has a modulus of 1 and an angle of \(\frac{\pi}{2}\). We need \(i^x = 1\). In polar terms: \(e^{i x \frac{\pi}{2}} = e^{i2\pi k}\) where \(k\) is an integer, this leads to \(x \frac{\pi}{2} = 2\pi k\).

Step 3: Solve the equation for x

From the previous step, we equate \(x \frac{\pi}{2} = 2\pi k\). Dividing both sides by \(\frac{\pi}{2}\) gives \(x = 4k\). Hence, \(x\) must be a multiple of 4.

Key Concepts

Polar Form

Complex numbers can be expressed in several different forms. One powerful form is the polar form. The polar form uses the magnitude of the complex number and the angle it forms with the positive real axis. For a complex number, like our exercise's final simplification of \(rac{1+i}{1-i} = i\), finding its polar form can simplify computations drastically.
In polar form, a complex number \(z = x + yi\) is expressed as \(re^{i\theta}\) where:

• \(r\) is the magnitude or modulus, \(r = \sqrt{x^2 + y^2}\).
• \(\theta\) is the argument, the angle calculated by \(\tan^{-1}(\frac{y}{x})\).

For the imaginary unit \(i\), the modulus is 1 (since \(\sqrt{0^2 + 1^2} = 1\)), and the argument is \(\frac{\pi}{2}\) because it lies on the positive imaginary axis. Therefore, we can write \(i\) as \(e^{i\frac{\pi}{2}}\). Any calculations involving powers or roots become a matter of simple exponent manipulation when using this form.

Complex Conjugate

The concept of a complex conjugate is quite useful when dealing with complex numbers. It helps to rationalize denominators among other purposes. A complex conjugate of a number \(z = a + bi\) is denoted \(\overline{z}\) and is defined as \(a - bi\). Conjugates have the property \(z \cdot \overline{z} = a^2 + b^2\), a real number.
In our exercise, the complex conjugate was used to simplify the fraction \(\frac{1+i}{1-i}\). By multiplying both the numerator and the denominator by the conjugate of the denominator \(1+i\), the expression becomes easier to handle, reducing to pure imaginary form. This contributes importantly to expressing it in polar form, since the result \(i\) is ready to be converted directly to \(e^{i\frac{\pi}{2}}\). This shows how complex conjugates are used to clear up expressions in complex calculations.

Roots of Unity

The roots of unity are particular solutions of the equation \(z^n = 1\) for complex numbers \(z\). These roots are evenly spaced around the unit circle in the complex plane, and their understanding is vital in fields like signal processing and algebra.
In our exercise, we need to solve the equation \(i^x = 1\). In polar terms, this translates to finding \(x\) such that \(e^{ix\frac{\pi}{2}} = e^{i2\pi k}\) where \(k\) is some integer. Effectively, this is asking us to identify powers of \(i\) that return to 1, akin to finding roots of unity.
The calculations show that thus \(x\) must be a multiple of 4, which aligns with the 4th roots of unity. When \(x\) takes multiples of 4, the angle \(x\frac{\pi}{2}\) becomes an integer multiple of \(2\pi\), completing a full rotation around the unit circle and returning to 1 on the real axis.