Answered

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find the cubed roots of 8(cos216 + i sin216)

Sagot :

marianamares2020

Step-by-step explanation:

z

3

=8(cos216

∘

+isin216

∘

)

z^3=2^3(\cos(6^3)^\circ+i\sin(6^3)^\circ)z

3

=2

3

(cos(6

3

)

∘

+isin(6

3

)

∘

)

\implies z=8^{1/3}\left(\cos\left(\dfrac{216+360k}3\right)^\circ+i\sin\left(\dfrac{216+360k}3\right)^\circ\right)⟹z=8

1/3

(cos(

3

216+360k

)

∘

+isin(

3

216+360k

)

∘

)

where k=0,1,2k=0,1,2 . So the third roots are

\begin{gathered}z=\begin{cases}2(\cos72^\circ+i\sin72^\circ)\\2(\cos192^\circ+i\sin192^\circ)\\2(\cos312^\circ+i\sin312^\circ)\end{cases}\end{gathered}

z=

⎩

⎪

⎪

⎨

⎪

⎪

⎧

2(cos72

∘

+isin72

∘

)

2(cos192

∘

+isin192

∘

)

2(cos312

∘

+isin312

∘

)

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