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Find the cube roots of 8(cos pi/3 + i sin pi/3)?
5 Answers
- Anonymous5 years ago
Let r(cos x + i sin x ) be a cube root of 8(cos pi/3 + i sin pi/3) then r^3 (cos 3x + i sin 3x ) = 8(cos pi/3 + i sin pi/3) Equating modulus and argument we have r^3 = 8 so r = 2 and 3x = pi/3 + 2kpi Where k is any integer. Hence x = pi/9 + 2kpi/3, and k = 0,1,2 gives distinct values. So the 3 cube roots of 8(cos pi/3 + i sin pi/3) are 2( cos x +isinx) where x is given above.
- Anonymous1 decade ago
I don't know how to do that. I suck at math. That is math. Right?
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