Answer to Question #339520 in Complex Analysis for Momo

Question #339520

3. 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
2022-05-12T09:23:25-0400

At first, we rewrite "z=1+i\\sqrt{3}" in the polar form. We get: "z=2(\\frac12+i\\frac{\\sqrt{3}}2)=2(\\cos(\\frac{\\pi}{3})+i\\,\\sin(\\frac{\\pi}{3}))".

We use the known formula (extension of de Moivre's formula) to compute the root "\\frac13". We get:

"z^{\\frac13}=2^{\\frac13}(\\cos(\\frac{\\frac{\\pi}3+2\\pi k}{3})+i\\,\\sin(\\frac{\\frac{\\pi}3+2\\pi k}{3}))," where "k=0,1,2."

We receive the following roots:

"v_0=2^{\\frac13}(\\cos(\\frac{{\\pi}}{9})+i\\,\\sin(\\frac{{\\pi}}{9}))",

"v_1=2^{\\frac13}(\\cos(\\frac{{7\\pi}}{9})+i\\,\\sin(\\frac{{7\\pi}}{9})),"

"v_2=2^{\\frac13}(-\\cos(\\frac{{4\\pi}}{9})-i\\,\\sin(\\frac{{4\\pi}}{9}))".

Answer: the cube root of "z" has the following values:

"v_0=2^{\\frac13}(\\cos(\\frac{{\\pi}}{9})+i\\,\\sin(\\frac{{\\pi}}{9}))",

"v_1=2^{\\frac13}(\\cos(\\frac{{7\\pi}}{9})+i\\,\\sin(\\frac{{7\\pi}}{9})),"

"v_2=2^{\\frac13}(-\\cos(\\frac{{4\\pi}}{9})-i\\,\\sin(\\frac{{4\\pi}}{9}))".


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