Question

Solve this !!
Confused!!!!!!!!!​

Answer 1

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

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

$\large\underline{\sf{To\:Find - }}$

The smallest positive integer n

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

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

can be rewritten as on Rationalizing,

$\rm :\longmapsto\: {\bigg(\dfrac{1 - i}{1 + i} \times \dfrac{1 - i}{1 - i} \bigg) }^{ {n}^{2} } = 1$

We know,

$\red{\bigg \{ \because \: (x + y)( x- y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm :\longmapsto\: {\bigg(\dfrac{ {(1 - i)}^{2} }{ {1}^{2} - {i}^{2} } \bigg) }^{ {n}^{2} } = 1$

We know that,

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

and

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

So, using all these Identities, we get

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

$\rm :\longmapsto\: {\bigg(\dfrac{1 - 1 - 2i}{1 - ( - 1) } \bigg) }^{ {n}^{2} } = 1$

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

$\rm :\longmapsto\: {\bigg(\dfrac{- 2i}{2} \bigg) }^{ {n}^{2} } = 1$

$\rm :\longmapsto\: {\bigg({- i}{} \bigg) }^{ {n}^{2} } = 1$

Now, we know that,

$\boxed{ \rm{ {i}^{4} = 1}}$

So, it means,

$\rm :\longmapsto\: {\bigg({- i}{} \bigg) }^{ {n}^{2} } = {i}^{4}$

can be further rewritten as

$\rm :\longmapsto\: {\bigg({- i}{} \bigg) }^{ {n}^{2} } = {( - i)}^{4}$

$\rm :\implies\: {n}^{2} = 4$

$\bf\implies \:n = 2$

Additional Information :-

$\boxed{ \rm{ {i}^{2} = - 1}}$

$\boxed{ \rm{ {i}^{3} = - i}}$

$\boxed{ \rm{ {i}^{4n} = 1 \: where \: n \: is \: natural \: number}}$

If n is a natural number, then

$\boxed{ \rm{ {i}^{n} + {i}^{n + 1} + {i}^{n + 2} + {i}^{n + 3} = 0}}$

$\boxed{ \rm{ \frac{1}{i} = - \: i}}$