Question #275261

Find z1/n; for n=3, z = 1-i in C (, the Argand Plane).

1
Expert's answer
2021-12-06T16:25:58-0500

The polar form of 1i=2(cos(π4+isin(π4)).1−i=\sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4})).

According to the De Moivre's Formula, all 3-th roots of a complex number 

2(cos(π4+isin(π4))\sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4})) are given by


(2)1/3(cos(π/4+2πk3)+isin(π/4+2πk3)),(\sqrt{2})^{1/3}(\cos(\dfrac{-\pi/4+2\pi k}{3})+i\sin(\dfrac{-\pi/4+2\pi k}{3})),k=0,1,2k=0,1,2

k=0:k=0:


(2)1/6(cos(π/4+2π(0)3)+isin(π/4+2π(0)3))(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (0)}{3})+i\sin(\dfrac{-\pi/4+2\pi (0)}{3}))

=(2)1/6(cos(π12)isin(π12))=(2)^{1/6}(\cos(\dfrac{\pi}{12})-i\sin(\dfrac{\pi}{12}))

k=1:k=1:


(2)1/6(cos(π/4+2π(1)3)+isin(π/4+2π(1)3))(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (1)}{3})+i\sin(\dfrac{-\pi/4+2\pi (1)}{3}))

=(2)1/6(cos(7π12)+isin(7π12))=(2)^{1/6}(\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12}))

k=2:k=2:


(2)1/6(cos(π/4+2π(2)3)+isin(π/4+2π(2)3))(2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (2)}{3})+i\sin(\dfrac{-\pi/4+2\pi (2)}{3}))

=(2)1/6(cos(5π4)+isin(5π4))=(2)^{1/6}(\cos(\dfrac{5\pi}{4})+i\sin(\dfrac{5\pi}{4}))

=(2)1/3(1+i)=(2)^{-1/3}(1+i)


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