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+i3z=1+i\sqrt{3} in the polar form. We get: z=2(12+i32)=2(cos(π3)+isin(π3))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 13\frac13. We get:

z13=213(cos(π3+2πk3)+isin(π3+2πk3)),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.k=0,1,2.

We receive the following roots:

v0=213(cos(π9)+isin(π9))v_0=2^{\frac13}(\cos(\frac{{\pi}}{9})+i\,\sin(\frac{{\pi}}{9})),

v1=213(cos(7π9)+isin(7π9)),v_1=2^{\frac13}(\cos(\frac{{7\pi}}{9})+i\,\sin(\frac{{7\pi}}{9})),

v2=213(cos(4π9)isin(4π9))v_2=2^{\frac13}(-\cos(\frac{{4\pi}}{9})-i\,\sin(\frac{{4\pi}}{9})).

Answer: the cube root of zz has the following values:

v0=213(cos(π9)+isin(π9))v_0=2^{\frac13}(\cos(\frac{{\pi}}{9})+i\,\sin(\frac{{\pi}}{9})),

v1=213(cos(7π9)+isin(7π9)),v_1=2^{\frac13}(\cos(\frac{{7\pi}}{9})+i\,\sin(\frac{{7\pi}}{9})),

v2=213(cos(4π9)isin(4π9))v_2=2^{\frac13}(-\cos(\frac{{4\pi}}{9})-i\,\sin(\frac{{4\pi}}{9})).


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