Question #336656

Use De Moivre’s Theorem to determine the cube root of Z and leave your answer in polar


form with the angle in radians


(a) Z = 1+i√3


1
Expert's answer
2022-05-03T16:44:47-0400

a) The polar form of 1+i31+i\sqrt{3} is

2(cos(π3)+isin(π3))2\big(\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})\big)

We have that r=2,θ=π3,n=3.r=2, \theta=\dfrac{\pi}{3}, n=3.

According to the De Moivre's Formula

k=0:k=0:


23(cos(π/3+2π(0)3)+isin(π/3+2π(0)3))\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(0)}{3})+i\sin(\dfrac{\pi/3+2\pi(0)}{3})\big)


=23(cos(π9)+isin(π9))=\sqrt[3]{2}\big(\cos(\dfrac{\pi}{9})+i\sin(\dfrac{\pi}{9})\big)

k=1:k=1:


23(cos(π/3+2π(1)3)+isin(π/3+2π(1)3))\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(1)}{3})+i\sin(\dfrac{\pi/3+2\pi(1)}{3})\big)


=23(cos(7π9)+isin(7π9))=\sqrt[3]{2}\big(\cos(\dfrac{7\pi}{9})+i\sin(\dfrac{7\pi}{9})\big)

=23(cos(π9)+isin(π9))=\sqrt[3]{2}\big(-\cos(\dfrac{\pi}{9})+i\sin(\dfrac{\pi}{9})\big)

k=2:k=2:


23(cos(π/3+2π(2)3)+isin(π/3+2π(2)3))\sqrt[3]{2}\big(\cos(\dfrac{\pi/3+2\pi(2)}{3})+i\sin(\dfrac{\pi/3+2\pi(2)}{3})\big)


=23(cos(13π9)+isin(13π9))=\sqrt[3]{2}\big(\cos(\dfrac{13\pi}{9})+i\sin(\dfrac{13\pi}{9})\big)

=23(cos(4π9)isin(4π9))=\sqrt[3]{2}\big(-\cos(\dfrac{4\pi}{9})-i\sin(\dfrac{4\pi}{9})\big)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS