Use the equation: r^(p/q){cos[p(theta +2n(pi))/q] +isin[p(theta + 2n(pi))/q]}, n

Question

Use the equation: r^(p/q){cos[p(theta +2n(pi))/q] +isin[p(theta + 2n(pi))/q]},

n= 0,1,...q-1

Use the above equation to compute all rational roots listed:


e) (-4)^(2/5)

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Explanation / Answer

(e) (-4 = 4 e^{i(2n+1)pi}) ((-4)^{2/5} = 16^{1/5} e^{i2(2n+1)pi/5}) n=0 => (16^{1/5} e^{i2pi /5} = 16^{1/5} (cos2pi/5 + i sin2pi/5)) n=1 => (16^{1/5} e^{i(pi /5+ pi)} = -16^{1/5} e^{ipi /5} = -16^{1/5} (cos2pi/5 + i sin2pi/5)) n=2 => (16^{1/5} e^{i2pi} = 16^{1/5}) n=3 => (16^{1/5} e^{i14pi/5} =16^{1/5} e^{i(15-1)pi/5} = - 16^{1/5} e^{-ipi/5} =-16^{1/5} (cospi/5 - i sinpi/5)) n=4 => (16^{1/5} e^{i18pi/5} = 16^{1/5} e^{i(20-2)pi/5} = 16^{1/5} e^{-i2pi/5} = 16^{1/5} (cos2pi/5 - i sin2pi/5))

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