Posted by **Kate** on Friday, May 17, 2013 at 5:01pm.

Can someone please give me a step by step answer on how to find the complex cube roots of 8(cos(4π/5)+isin(4π/5))? I need to know how to do this for my trig final next week.

- TRIGONOMETRY -
**Reiny**, Friday, May 17, 2013 at 5:52pm
We will use De Moivre's theorem

8(cos(4π/5) + isin(4π/5)^(1/3)

= 8^(1/3) (cos (1/3)(4π/5) + i sin (1/3)(4π/5) )

= 2(cos (4π/15) + i sin(4π/15) )

for a youtube of De Moivre's theorem check out this:

http://www.youtube.com/watch?v=NrwWv9JHdI4

- TRIGONOMETRY -
**Steve**, Friday, May 17, 2013 at 5:57pm
(rcisθ)^n = r^n cis(nθ)

cuberoot = 1/3 power, so

∛8(cos(4π/5)+isin(4π/5))

= ∛8 cis((4π/5)/3)

= 2cis(4π/15)

Now, since cisθ = cis(θ+2nπ)

cis 4π/5 = cis 14π/5 = cis 24π/5

(2cis 4π/15)^3 = 8cis 4π/5

(2cis (4π/15 + 2π/3))^3 = (2cis 14π/15)^3 = 8cis 14π/5

(2cis (4π/15 + 4π/3))^3 = (2cis 24π/15)^3 = 8cis 24π/5

So, you need to evaluate sin & cos of 4π/15,14π/15,24π/15

adding another 2π takes us around to 34π/15 = 2π + 4π/15 and we're back where we started.

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