Problem 85 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The integral solution of the equation \((1-i)^{n}=2^{n}\) is (A) \(n=0\) (B) \(n=1\) (C) \(n=-1\) (D) None of these

Step-by-Step Solution

Answer

The integral solution is \(n=0\).

Step 1: Understanding Complex Numbers

We need to solve \((1-i)^n = 2^n\).Systematically, each side of the equation is a complex number raised to a power. The left side contains the complex number \((1-i)\), while the right is \(2\), a real number.

Step 2: Polar Form Conversion

Convert \(1-i\) to polar form. Determine its modulus:\[|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\].Find the argument (angle) of the complex number:\[\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}\].Thus, in polar form: \((1-i) = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)\)

Step 3: Apply De Moivre's Theorem

Use De Moivre's Theorem to raise \((1-i)\) to the power of \(n\):\[(1-i)^n = \left(\sqrt{2}\right)^n \left(\cos\left(n \cdot -\frac{\pi}{4}\right) + i \sin\left(n \cdot -\frac{\pi}{4}\right)\right)\]Compute the expression:\[2^{n/2} \left(\cos\left(-\frac{n\pi}{4}\right) + i \sin\left(-\frac{n\pi}{4}\right)\right)\].

Step 4: Equate Magnitude and Argument

Since \[(1-i)^n = 2^n\],we equate the magnitudes:\[2^{n/2} = 2^n\].Solving for \(n\), we get:\[n/2 = n\] implying \(n = 0\) as a potential solution.For arguments, since the right side \(2^n\) does not have a complex angle, the only valid angle for the left side is a multiple of \(2\pi\). This confirms \(\frac{-n\pi}{4} = 2k\pi\) holds true for \(n = 0\) where \(k\) is an integer.

Step 5: Verify Solution

We verified both modulus and argument match if \(n = 0\). By testing values and restoring the identity, it's confirmed that the equation holds accurate for \(n=0\) as no other integer satisfies both conditions.

Key Concepts

Polar Form Conversion

Complex numbers can be represented in different forms. While this might initially seem daunting, it greatly simplifies operations like multiplication and exponentiation. The polar form of a complex number takes advantage of trigonometric functions.When converting a complex number like \(1-i\) into polar form, we begin by finding its modulus. The modulus is the distance from the origin in the complex plane, calculated as:

• \(|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)

Next, we calculate the argument, which is the angle the complex number makes with the positive real axis. It is given by:

• \(\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}\)

Therefore, the polar form of the complex number \(1-i\) becomes:

• \(1-i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)\)

This conversion transforms the complex number into a format ideal for using other mathematical theorems. It also gives a clearer geometric interpretation of complex numbers.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool that makes working with powers of complex numbers much simpler. It states that for a complex number in polar form, raising it to an integer power involves using its modulus and multiplying the argument by the power.The theorem can be stated as:

• If \(z = r\left(\cos\theta + i\sin\theta\right)\)
• Then \(z^n = r^n \left(\cos\left(n\theta\right) + i\sin\left(n\theta\right)\right)\)

For our problem, we apply this theorem to \((1-i)^n\), giving:

• \((1-i)^n = \left(\sqrt{2}\right)^n \left(\cos\left(n\cdot -\frac{\pi}{4}\right) + i \sin\left(n\cdot -\frac{\pi}{4}\right)\right)\)
• This simplifies to \(2^{n/2}\) times the trigonometric expression.

This representation allows for easy comparison with real numbers like \(2^n\). De Moivre's provides a structured method to dissect and solve complex exponential equations by focusing individually on magnitude and argument.

Integral Solutions

Finding integral solutions to equations involving complex numbers can involve confirming both magnitude and angle consistency. For the equation \((1-i)^n = 2^n\), we need integral \(n\) that satisfies both:

• \(\left| (1-i)^n \right| = \left| 2^n \right|\)
• The argument of \((1-i)^n\) is compatible with a real number like \(2^n\), which has zero argument.

First, equating the magnitudes gives:

• \(2^{n/2} = 2^n\), implying \(n/2 = n\) or \(n = 0\).

The argument condition (\(\frac{-n\pi}{4} = 2k\pi\) for some integer \(k\)) also confirms \(n = 0\), as only this \(n\) yields an argument of zero.Through this step-by-step verification, it's reinforced that the only integral value for which both magnitude and argument align is \(n=0\). This ensures no other integer values satisfy the equation, making \(n=0\) the sole solution.