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28 August, 12:30

Find the fifth roots of 243 (cos 260° + i sin 260°)

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  1. 28 August, 14:12
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    use De Moivre's Theorem:

    ⁵√[243 (cos 260° + i sin 260°) ] = [243 (cos 260° + i sin 260°) ]^ (1/5)

    = 243^ (1/5) (cos (260 / 5) ° + i sin (260 / 5) °)

    = 3 (cos 52° + i sin 52°)

    z1 = 3 (cos 52° + i sin 52°) ←← so that's the first root

    there are 5 roots so the angle between each root is 360/5 = 72°

    then the other four roots are:

    z2 = 3 (cos (52 + 72) ° + i sin (52 + 72) °) = 3 (cos 124° + i sin 124°)

    z3 = 3 (cos (124 + 72) ° + i sin (124 + 72) °) = 3 (cos 196° + i sin 196°)

    z4 = 3 (cos (196 + 72) ² + i sin (196 + 72) °) = 3 (cos 268° + i sin 268°)

    z5 = 3 (cos (268 + 72) ° + i sin (268 + 72) °) = 3 (cos 340° + i sin 340°)
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