If $\left(\dfrac{1+i}{1-i}\right)^x=1$, then: (A) $x=2n+1$ (B) $x=4n$ (C) $x=2n$ (D) $x=4n+1$, where $n\in\mathbf{N}$

> [!TLDR] > The complex fraction simplifies to $i$, so the problem becomes finding $x$ such that $i^x=1$. This occurs when $x$ is a multiple of 4, meaning $\boxed{x=4n}$. When dealing with powers of complex numbers, especially fractions involving $i$, the first step is almost always to simplify the base of the power. This often involves rationalizing the denominator or converting to polar form. In this problem, the expression $\left(\frac{1+i}{1-i}\right)$ is a classic form that simplifies beautifully. Understanding this simplification is key to solving the problem efficiently. The core idea is to transform the complex number inside the parenthesis into its simplest form, then use the cyclic properties of powers of $i$ to determine the possible values of $x$. 1. **Simplify the complex fraction:** We begin by simplifying the base of the power, $\frac{1+i}{1-i}$. To do this, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $1-i$ is $1+i$. $$ \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} $$ Now, we expand the numerator and the denominator. For the numerator: $(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$. For the denominator: $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$. Substituting these back into the fraction: $$ \frac{1+i}{1-i} = \frac{2i}{2} = i $$ > [!TIP] > The expression $\frac{1+i}{1-i}$ frequently appears in complex number problems. It's worth remembering that it simplifies directly to $i$. Similarly, $\frac{1-i}{1+i}$ simplifies to $-i$. 2. **Substitute the simplified base into the equation:** Now that we've simplified the base, the original equation $\left(\frac{1+i}{1-i}\right)^x=1$ transforms into: $$ i^x = 1 $$ 3. **Determine the values of $x$ for which $i^x=1$:** We need to recall the cyclic nature of powers of $i$: * $i^1 = i$ * $i^2 = -1$ * $i^3 = i^2 \cdot i = -1 \cdot i = -i$ * $i^4 =

If ((1+i)/(1-i))x=1, then: (A) x=2n+1 (B) x=4n (C) x=2n (D) x=4n+1, where n∈mathbfN

NCERT Exemplar · Q27

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

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Concept understanding — Complex Number Transform

The Intuition: Why "Transform" a Number?

You already know that a real number lives on a line — the number line. Adding $+2$ slides you right; multiplying by $-1$ flips you to the opposite side. But what if you want to rotate something? A real number can't do that on its own. Multiplying by $-1$ is a 180° rotation, but what about a 90° rotation? That's where the complex number transform comes in.

Think of a complex number $z = a + bi$ as a point (or an arrow) on a 2D plane. The real part $a$ is the horizontal coordinate, the imaginary part $b$ is the vertical coordinate. Now, when you multiply two complex numbers, something beautiful happens: the lengths multiply, and the angles add.

Note

This is the core insight: multiplication of complex numbers is a rotation + scaling operation, not just a scaling like real numbers.

So a "complex number transform" is simply the act of applying a complex number (as an operator) to another complex number (as a point) — usually by multiplication — to achieve a geometric transformation: rotation, scaling, or both.

The Precise Statement

Let $z = x + yi$ be any complex number (the "point" you want to transform).

Let $w = r(\cos\theta + i\sin\theta)$ be a fixed complex number (the "transformer").

Then the complex number transform of $z$ by $w$ is:

w \cdot z = r(\cos\theta + i\sin\theta) \cdot (x + yi)

When you multiply this out (using $i^2 = -1$ ), the result is a new complex number $z'$ whose geometric meaning is:

Scale the distance of $z$ from the origin by a factor of $r$
Rotate the point $z$ around the origin by an angle $\theta$ counterclockwise

\text{If } w = re^{i\theta}, \text{ then } w \cdot z \text{ rotates } z \text{ by } \theta \text{ and scales it by } r.

This is often written using Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$ , so $w = re^{i\theta}$ .

A Concrete Example

Take the point $z = 1 + 0i$ (the number 1 on the real axis).

Let the transformer be $w = i$ (which has $r=1$ , $\theta = 90^\circ$ ).

i \cdot 1 = i

The point $(1,0)$ moved to $(0,1)$ — a 90° rotation counterclockwise. No scaling because $|i| = 1$ .

Now take $z = 2 + 0i$ and $w = 2i$ (which has $r=2$ , $\theta = 90^\circ$ ):

2i \cdot 2 = 4i

The point $(2,0)$ moved to $(0,4)$ — rotated 90° and scaled by factor 2.

Watch out

A common mistake: thinking that multiplying by $i$ always gives a 90° rotation. It does — but only if you multiply the entire complex number. Multiplying just the real part by $i$ is not the same as multiplying the whole number.

Why This Matters

This transform is the foundation of: …

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The complex fraction simplifies to $i$ , so the problem becomes finding $x$ such that $i^x=1$ . This occurs when $x$ is a multiple of 4, meaning $\boxed{x=4n}$ .

When dealing with powers of complex numbers, especially fractions involving $i$ , the first step is almost always to simplify the base of the power. This often involves rationalizing the denominator or converting to polar form. In this problem, the expression $\left(\frac{1+i}{1-i}\right)$ is a classic form that simplifies beautifully. Understanding this simplification is key to solving the problem efficiently.

The core idea is to transform the complex number inside the parenthesis into its simplest form, then use the cyclic properties of powers of $i$ to determine the possible values of $x$ .

Simplify the complex fraction: We begin by simplifying the base of the power, $\frac{1+i}{1-i}$ . To do this, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $1-i$ is $1+i$ .

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

Now, we expand the numerator and the denominator.
For the numerator: $(1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$.
For the denominator: $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$.

Substituting these back into the fraction:

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

> [!TIP]
> The expression $\frac{1+i}{1-i}$ frequently appears in complex number problems. It's worth remembering that it simplifies directly to $i$. Similarly, $\frac{1-i}{1+i}$ simplifies to $-i$.

2. Substitute the simplified base into the equation:

Now that we've simplified the base, the original equation $\left(\frac{1+i}{1-i}\right)^x=1$ transforms into:

i^x = 1

Determine the values of $x$ for which $i^x=1$ : We need to recall the cyclic nature of powers of $i$ $i$ :
- $i^1 = i$
- $i^2 = -1$
- $i^3 = i^2 \cdot i = -1 \cdot i = -i$
- $i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1$
- $i^5 = i^4 \cdot i = 1 \cdot i = i$ (The cycle repeats) …

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Q.If (1+i1−i)x=1\left(\dfrac{1+i}{1-i}\right)^x=1(1−i1+i​)x=1, then: (A) x=2n+1x=2n+1x=2n+1 (B) x=4nx=4nx=4n (C) x=2nx=2nx=2n (D) x=4n+1x=4n+1x=4n+1, where n∈Nn\in\mathbf{N}n∈N

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Q.If $\left(\dfrac{1+i}{1-i}\right)^x=1$ , then:
(A) $x=2n+1$
(B) $x=4n$
(C) $x=2n$
(D) $x=4n+1$ , where $n\in\mathbf{N}$

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