Question

If i = StartRoot negative 1 EndRoot, what is the value of i 3?

Answer 1

-1

Explanation:

i = √-1 => i^2 = -1

i^3 = i^2 * i = -1 * i = -1

Answer 2

i³ = - i

Explanation:

Explanation:-

Step(i):-

Given
`i = √(-1)`

we know that cos π = -1

sin π = 0

now given data

we can write
`i = √(cos\pi + i sin\pi )`

`i =( {cos\pi + i sin\pi })^{(1)/(2) }`

step(ii):-

In complex numbers nth roots of a complex number

Z = r(cosθ+i sinθ)

`Z^{(n)/(2) } = r^{(n)/(2) } (cos\pi +isin\pi )^{(n)/(2) }`

`Z^{(n)/(2) } = r^{(n)/(2) } (cosn\pi/2 }+isinn\pi/2 )`

Step(iii):-

now
`i =( {cos(\pi )/(2) + i sin(\pi )/(2) }) }`

`i^(3) =( {cos(\pi )/(2) + i sin(\pi )/(2) })^(3) }`

`i^(3) =( {cos(3\pi )/(2) + i sin(3\pi )/(2) })}`

By using trigonometric

`cos(3\pi )/(2) = cos (\pi +(\pi )/(2) ) = -cos(\pi )/(2) =0`

`sin(3\pi )/(2) = sin (\pi +(\pi )/(2) ) = -sin(\pi )/(2) = -1`

`i^(3) =( 0+ i (-1) ) = -i`

Final answer:-

i³ = - i