Question

Which expression is a cube root of -1+i sqrt3

Answer 1

(cos(40) + isin(40))

(cos(120) + isin(120))

(cos(280) + isin(280))

(cos(320) + isin(320))

Answer 2

\[\sqrt{x^2+y^2} e ^{\tan^{-1} y/x}\]

Answer 3

None of the expressions given is the solution, the reason being that when \[-1+i \sqrt{3}\]
is converted to polar form r = 2. But when the cube root of the complex number is taken r will become
\[\sqrt[3]{2}\]
which is not in any of the choices.

Answer 4

z=-1+i sqrt(3)
||z||=sqrt(1+3)=2
z=2 (-0.5+i sqrt(3)/2)
cos(theta)=-0.5 .............. sin(theta)=sqrt(3)/2 -------> theta=120
z=2 exp(120 i )
\[\sqrt[3]{z}=\sqrt[3]{2}*e ^{(120/3+2k pi)i}\]
k=0,1,2
then
\[\sqrt[3]{z}=\sqrt[3]{2}*e ^{40i}=\sqrt[3]{2}*(\cos 40+ i \sin 40)\]