Answer:
i³ = - i
Step-by-step explanation:
Explanation:-
Step(i):-
Given [tex]i = \sqrt{-1}[/tex]
we know that cos π = -1
sin π = 0
now given data
we can write [tex]i = \sqrt{cos\pi + i sin\pi }[/tex]
[tex]i =( {cos\pi + i sin\pi })^{\frac{1}{2} }[/tex]
step(ii):-
In complex numbers nth roots of a complex number
Z = r(cosθ+i sinθ)
[tex]Z^{\frac{n}{2} } = r^{\frac{n}{2} } (cos\pi +isin\pi )^{\frac{n}{2} }[/tex]
[tex]Z^{\frac{n}{2} } = r^{\frac{n}{2} } (cosn\pi/2 }+isinn\pi/2 )[/tex]
Step(iii):-
now [tex]i =( {cos\frac{\pi }{2} + i sin\frac{\pi }{2} }) }[/tex]
[tex]i^{3} =( {cos\frac{\pi }{2} + i sin\frac{\pi }{2} })^{3} }[/tex]
[tex]i^{3} =( {cos\frac{3\pi }{2} + i sin\frac{3\pi }{2} })}[/tex]
By using trigonometric
[tex]cos\frac{3\pi }{2} = cos (\pi +\frac{\pi }{2} ) = -cos\frac{\pi }{2} =0[/tex]
[tex]sin\frac{3\pi }{2} = sin (\pi +\frac{\pi }{2} ) = -sin\frac{\pi }{2} = -1[/tex]
[tex]i^{3} =( 0+ i (-1) ) = -i[/tex]
Final answer:-
i³ = - i
