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Find the fifth roots of 32(cos 280° + i sin 280°). 0 votescalculusaskedAug 4, 2014in CALCULUSby Tdog79Pupil

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Let, z = 32(cos 2800 + i sin 2800) = 32 cos(3600*n + 2800) + i sin(3600*n + 2800), wright here n is an integer.

⇒ z1/5 = < 32 cos(3600*n + 2800) + i sin(3600*n + 2800) >1/5

From De - Moievere"s theorem.

z1/5 = 2 +

⇒ z1/5 = 2 cos(72n + 56) + i sin(72n + 56) , where n is an integer.

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