Question

If \((\frac{1+i}{1-i})^x=1\) then

• A. x = 4n + 1; n ∈ N
• B. x = 2n + 1, n ∈ N
• C. x = 2n; n ∈ N
• D. x = 4n; n ∈ N

Answer 1

The Correct Option is D

We are asked to solve the equation \( \left(\frac{1+i}{1-i}\right)^x = 1 \).

Step 1: Simplify the base \( \frac{1+i}{1-i} \) 

To simplify the fraction, we multiply the numerator and the denominator by the complex conjugate of the denominator, which is \( 1+i \):

\[ \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} \]

Calculate the numerator:

\[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]

Calculate the denominator:

\[ (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

So, the fraction simplifies to:

\[ \frac{1+i}{1-i} = \frac{2i}{2} = i \]

Step 2: Rewrite the equation

Substituting the simplified base back into the original equation, we get:

\[ i^x = 1 \]

Step 3: Solve for x

We need to find the values of x for which \( i^x = 1 \).

Recall the powers of \( i \):

• \( i^1 = i \)
• \( i^2 = -1 \)
• \( i^3 = i^2 \cdot i = -i \)
• \( i^4 = i^2 \cdot i^2 = (-1)(-1) = 1 \)
• \( i^5 = i^4 \cdot i = 1 \cdot i = i \)
• \( i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1 \)
• \( i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i \)
• \( i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1 \)

The powers of \( i \) repeat in a cycle of 4. We see that \( i^x = 1 \) whenever x is a multiple of 4.

Therefore, the general solution for integer x is \( x = 4k \), where \( k \) is any integer (\( k \in \mathbb{Z} \)).

Step 4: Compare with the given options

The options provide expressions for x in terms of n, where \( n \in N \) (natural numbers, typically \( n = 1, 2, 3, \dots \)).

• x = 4n + 1; n ∈ N (e.g., 5, 9, 13, ...) → \( i^x = i \)
• x = 2n + 1, n ∈ N (e.g., 3, 5, 7, ...) → \( i^x = -i \) or \( i \)
• x = 2n; n ∈ N (e.g., 2, 4, 6, 8, ...) → \( i^x = -1 \) or \( 1 \)
• x = 4n; n ∈ N (e.g., 4, 8, 12, ...) → \( i^x = 1 \)

The option \( x = 4n; n \in N \) correctly describes the positive integer values of x that are multiples of 4, satisfying the condition \( i^x = 1 \).

The correct option is x = 4n; n ∈ N.

Answer 2

The given equation is \( \left(\frac{1+i}{1-i}\right)^x = 1 \). We can simplify the fraction \( \frac{1+i}{1-i} \) by multiplying the numerator and denominator by \( 1+i \) to get: \[ \frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i \] Therefore, the equation becomes: \[ i^x = 1 \] The powers of \( i \) cycle every 4 terms as \( i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1 \).
Thus, for \( i^x = 1 \), \( x \) must be a multiple of 4, i.e., \( x = 4n \), where \( n \) is a natural number.

The correct answer is (D) : x = 4n; n ∈ N