Question

Answer this question pls. And show the detailed steps also. No spam answers.

Answer 1

Given Question :-

The smallest positive integer n for which

$\rm :\longmapsto\: {\bigg[\dfrac{1 + i}{1 - i} \bigg]}^{n} = 1$

(a) 1

(b) 2

(c) 3

(d) 4

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: {\bigg[\dfrac{1 + i}{1 - i} \bigg]}^{n} = 1$

On rationalizing the denominator, we get

$\rm :\longmapsto\: {\bigg[\dfrac{1 + i}{1 - i} \times \dfrac{1 + i}{1 + i} \bigg]}^{n} = 1$

We know,

$\boxed{ \tt{ \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \: }}$

and

$\boxed{ \tt{ \: (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using these Identities, we get

$\rm :\longmapsto\: {\bigg[\dfrac{ 1+ {i}^{2} + 2i }{ {1}^{2} - {i}^{2} } \bigg]}^{n} = 1$

We know,

$\boxed{ \tt{ \: {i}^{2} = - 1 \: }}$

So, using this, we get

$\rm :\longmapsto\: {\bigg[\dfrac{1 - 1 + 2i}{1 + 1} \bigg]}^{n} = 1$

$\rm :\longmapsto\: {\bigg[\dfrac{2i}{2} \bigg]}^{n} = 1$

$\rm :\longmapsto\: {i}^{n} = 1$

$\rm :\longmapsto\: {i}^{n} = {i}^{4}$

$\bf\implies \:\boxed{ \tt{ \: \: n \: = \: 4 \: \: }}$

• Hence, Option (d) is correct.

More to know :-

$\red{\rm :\longmapsto\:i = \sqrt{ - 1} \: }$

$\red{\rm :\longmapsto\: {i}^{3} = - 1}$

$\red{\rm :\longmapsto\: {i}^{4} = 1}$

$\red{\rm :\longmapsto\: {i}^{x} + {i}^{x + 1} + {i}^{x + 2} + {i}^{x + 3} = 0 \: where \: x \: is \: natural \: number}$

$\red{\rm :\longmapsto\:\dfrac{1}{i} = - i}$