If ((1+i)/(1-i))^x = 1 then (A) x = 2n + 1, where n is any positive integer (B) x = 4n, where n is a

Q: Wait, why do we multiply by $(1+i)$? Can't we just use polar form?

You absolutely can use polar form! Converting to $e^{i\theta}$ is a powerful method. However, for a simple fraction like $\frac{1+i}{1-i}$, rationalization is often faster and less prone to calculation errors. Both paths lead to the same result, $i$, but rationalization keeps us firmly in the algebraic domain.

Q: Why is the answer $x=4n$ and not just $x=4$?

Great question! The equation $i^x = 1$ is periodic. The powers of $i$ follow a cycle: $i^1=i, i^2=-1, i^3=-i, i^4=1, i^5=i$, and so on. Since $i^4=1$, any multiple of 4 will also result in 1 (e.g., $i^8 = (i^4)^2 = 1^2 = 1$). Thus, $x$ must be any multiple of 4, which we represent as $4n$.

Q: Is $x=0$ a valid solution?

Mathematically, $i^0 = 1$, so $x=0$ is a solution. However, in the context of JEE problems and the provided options, we are typically looking for positive integers $n$. The options specify $n$ as a positive integer, so we focus on $4, 8, 12, \dots$.

Question

If $\left(\frac{1+i}{1-i}\right)^x = 1$ then

Options:

(A) $x = 2n + 1$, where $n$ is any positive integer

(B) $x = 4n$, where $n$ is any positive integer

(C) $x = 2n$, where $n$ is any positive integer

(D) $x = 4n + 1$, where $n$ is any positive integer

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to solve for $x$ in the equation:

$$\left( \frac{1+i}{1-i} \right)^x = 1$$

At first glance, this appears to be a daunting task involving complex algebra. However, in JEE mathematics, complexity is often a mask for hidden simplicity.

### Simplifying the Base

Our first mission is to unmask the base of the exponent. We focus on the complex fraction:

$$\frac{1+i}{1-i}$$

To simplify this, we multiply both the numerator and the denominator by the complex conjugate of the denominator, which is $(1+i)$.

The denominator becomes:

$$(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2$$

The numerator becomes:

$$(1+i)(1+i) = (1+i)^2 = 1^2 + 2i + i^2 = 1 + 2i - 1 = 2i$$

Combining these results, the fraction simplifies beautifully:

$$\frac{2i}{2} = i$$

### The Cyclic Dance of $i$

The original equation has now transformed into the much friendlier form:

$$i^x = 1$$

To solve this, we visualize $i$ as a rotation operator on the complex plane. Multiplying by $i$ is equivalent to a $90^{\circ}$ counter-clockwise rotation.

Let us trace the cycle:

$i^1 = i$ (positive imaginary axis)

$i^2 = -1$ (negative real axis)

$i^3 = -i$ (negative imaginary axis)

$i^4 = 1$ (positive real axis)

### Final Conclusion

We have completed a full circle, and this cycle repeats every four steps. For $i^x$ to equal $1$, $x$ must land exactly on the positive real axis.

This condition is satisfied only when $x$ is a multiple of $4$. Therefore, the solution is:

**$x = 4n$, where $n \in \{1, 2, 3, \dots\}$**

You have successfully navigated the complexity, simplified the base, and utilized the cyclic nature of the imaginary unit to reach the elegant truth.

Animated Solution for Mathematics - Complex Numbers: If $(\frac{1 + i}{1 - i})^{x} = 1$ then

Select Answer:

Visualized Solution

The Complex Equation

Isolating the Base

Multiplying by the Conjugate

Expanding the Numerator

Simplifying the Numerator

Expanding the Denominator

Simplifying the Denominator

Final Simplification of the Base

The Simplified Equation

Powers of $i$ : $i^{1}$ and $i^{2}$

Powers of $i$ : $i^{3}$ and $i^{4}$

The Cyclic Nature of $i$

Final Solution

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Simplifying the Base

The Cyclic Dance of $i$

Final Conclusion

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (1−i1+i​)x=1 then