Question

Show that,
$( - \sqrt{ - 1} {)}^{4n + 3} = i \: where \: n \: is \: a \: positive \: integer$
​

Answer 1

⏩Hey mate

Show that

[-(√-1)]^(4n+3) = i,

where n is a positive integers

Proof:-

given here

=[-(√-1)]^(4n+3)

let

n= 0,1,2,3,4,...........

So,

if n=0,

so,

= [-(√-1)]^(0+3)

=[-(√-1)]^3

=(-)³ (√-1)³

= -[(√-1)(√-1)(√-1)]

= -[ (-1)(√-1)]

= √-1

=i

we know

i= √-1

Again, we take

if n= 1,

= [-(√-1)]^(4+3)

=[-(√-1)]^(7)

=(-1)^7 [(√-1)^7]

= - [ - (√-1)]

= √-1

= i

so,

we can take here n = 0,1,2,3,4,....

and show that

[-(√-1)]^(4n+3)= i

that's proved

⏩Hopes its helps u

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Answer 2

$\mathfrak{\underline{\underline{\large{Answer:-}}}}$

$\tt{\red{We \: have}}$

$\sf{( - \sqrt{ - 1} )^{4n + 3} = ( - i)^{4n + 3} }$

$\sf{ [ \sqrt{ - 1} = i] }$

$\sf{ = ( - i)^{4n} \times ( - i)^{3} }$

$\sf{ [ = ( - i {)}^{4} {]}^{n} \times ( - i {)}^{3} = (1 \times i) = i}$

$\sf{ [ ( - i {)}^{4} = 1 \: and \: - {i}^{3} = - ( - i) = i ] }$

$\tt{\red{Hence, ( - \sqrt{ - 1} {)}^{4n + 3} = i}}$