Answer to Question #336656 in Complex Analysis for Busi

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+i\\sqrt{3}" is

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

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

According to the De Moivre's Formula

"k=0:"


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


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

"k=1:"


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


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

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

"k=2:"


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


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

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


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