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:

a) sqrt (2i)

b) i^(1/3)

c) 1^(2/3)

d) (1+ i) ^(1/4)

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

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Explanation / Answer

(a) (sqrt{2i} = sqrt{2 e^{i(?/2+2n?)}}=sqrt{2} e^{i(?/4+n?)} n=0 => (sqrt{2} e^{ipi/4}=1+i ) n=1 => (sqrt{2} e^{i(?/4+?)}=?1?i ) (b)(i=e^{ipi/2+2npi}) (i^{1/3} = e^{i(1/2+2n)pi/3} = e^{i(1+4n)pi/6}) n=0 => (e^{ipi/6} = cospi/6 + i sinpi/6) n=1 => (e^{i5pi/6} = e^{-ipi/6} = cospi/6 - i sinpi/6) n=2 => (e^{i9pi/6} = - i) (c) (1=e^{i2npi}) (1^{1/3} = e^{i2npi/3}) n=0 => 1 n=1 => (e^{i2pi/3} = cos2pi/3 + i sin2pi/3) n=2 => (e^{i4pi/3} = - e^{ipi/3} = -cospi/3 - i sinpi/3) (d) (1+i = sqrt{2} e^{i(1/4 + 2n)pi} = sqrt{2} e^{i(8n+1)pi/4}) ((1+ i) ^(1/4) = 2^{1/8} e^{i(8n+1)pi/16}) n=0 => (2^{1/8} e^{ipi/16}) n=1 => (2^{1/8} e^{i9pi/16}) n=2 => (2^{1/8} e^{i(16+1)pi/16} =- 2^{1/8} e^{ipi/16}) n=3 => (2^{1/8} e^{i25pi/16} = 2^{1/8} e^{-i7pi/16}) (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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