Question #114823
9.3 Let w be a negative real number, z a 6
th root of w.
(a) Show that z (k) = ρ^
1
6
-

cos (pi+2kpi/6 )+ isin ( pi+2kpi/6), , k = 0, 1, 2, 3, 4, 5 is a formula for the
6th roots of w.
1
Expert's answer
2020-05-11T09:06:29-0400

(a) Let w=ρ(cosϕ+isinϕ)w=\rho(\cos\phi+i\sin\phi) and z=r(cosθ+isinθ)z=r(\cos\theta+i sin\theta) then by De Moivre's formula we have that zn=rn(cosnθ+isinnθ)z^n=r^n(\cos n\theta+i \sin n\theta)

zn=wrn(cosnθ+isinnθ)=ρ(cosϕ+isinϕ)z^n=w \rightsquigarrow r^n(\cos n\theta+i \sin n\theta)=\rho(\cos\phi+i\sin\phi)

Hence, rn=ρ,nθ=ϕ+2πkr^n=\rho, \,\, n\theta=\phi+2\pi k

wRϕ=πw\in\mathbb R_- \rightsquigarrow \phi=\pi

For n=6 ⁣:r=ρ1/6(cosπ+2πk6+isinπ+2πk6)n=6\colon r=\rho^{1/6} \bigg( \cos \frac{\pi+2\pi k}{6} +i\sin \frac{\pi+2\pi k}{6} \bigg)

for each k{0,1,,n1=5}k\in\{0,1,\dots,n-1=5\}




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