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Which expression is a cube root of -2i?

A. ^3rad2(cos(260 degrees)+isin(260 degrees))
B. ^3rad2(cos(60 degrees)+isin(60 degrees))
C. ^3rad2(cos(210 degrees)+isin(210 degrees))
D. ^3rad2(cos(180 degrees)+isin(180 degrees))

Respuesta :

mysticchacha

Answer:

for k = 0, 1, 2

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(90  + k*120)  + i sin (90 + k*120) )

Choice: C.  for k = 1

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(210)  + i sin (210) )

Step-by-step explanation:

Rewrite  -2i  as   2*( 0 - i) =  2 * (sin π +  i*cos π)

(-2i)^(1/3) =  [2 (cos 3π/2  + i sin 3π/2) ]^(1/3)

(-2i)^(1/3) =  2^(1/3)  (cos 3π/2  + i sin 3π/2)^(1/3)

(-2i)^(1/3) =  2^(1/3)  ( cos ((3π/2 + 2kπ)/3  )  + i sin ((3π/2 + 2kπ)/3 ) )

for k = 0, 1, 2

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos (π/2  + 2kπ/3)  + i sin (π/2 + 2kπ/3) )

for k = 0, 1, 2

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(90  + k*120)  + i sin (90 + k*120) )

for k = 0, 1, 2

so we can let k = 0  and get:

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(90)  + i sin (90) )

for k = 1

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(90  + 120)  + i sin (90 + 120) )

[tex]\sqrt[3]{-2i}[/tex] =  [tex]\sqrt[3]{2}[/tex]  ( cos(210)  + i sin (210) )

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