Question

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))

Answer 1

for k = 0, 1, 2

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

Choice: C. for k = 1

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

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

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

for k = 0, 1, 2

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

for k = 0, 1, 2

so we can let k = 0 and get:

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

for k = 1

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

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