Answer to Question #230184 in Complex Analysis for Anand

Question #230184

Say true or false, with a short proof.


All the cube roots of I in C are z₁ = cos(π/2)+ i sin(π/2) , z₂ = cos (π/6)+ isin(π/6) and z₃ = cos (5π/6)+i sin (5π/6)


1
Expert's answer
2021-08-29T16:40:55-0400

The polar form of ii is 1(cos(π2)+isin(π2)).1(\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})).

According to the De Moivre's Formula, all cubic roots of a complex number 1(cos(π2)+isin(π2))1(\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})) are given by 11/3(cos(π/2+2πk3)+isin(π/2+2πk3)),k=0,1,2.1^{1/3}(\cos(\dfrac{\pi/2+2\pi k}{3})+i\sin(\dfrac{\pi/2+2\pi k}{3})), k=0,1,2.


k=0,13(cos(π/2+2π(0)3)+isin(π/2+2π(0)3))k=0, \sqrt[3]{1}(\cos(\dfrac{\pi/2+2\pi (0)}{3})+i\sin(\dfrac{\pi/2+2\pi (0)}{3}))

=cos(π6)+isin(π6)=\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})



k=1,13(cos(π/2+2π(1)3)+isin(π/2+2π(1)3))k=1, \sqrt[3]{1}(\cos(\dfrac{\pi/2+2\pi (1)}{3})+i\sin(\dfrac{\pi/2+2\pi (1)}{3}))

=cos(5π6)+isin(5π6)=\cos(\dfrac{5\pi}{6})+i\sin(\dfrac{5\pi}{6})




k=2,13(cos(π/2+2π(2)3)+isin(π/2+2π(2)3))k=2, \sqrt[3]{1}(\cos(\dfrac{\pi/2+2\pi (2)}{3})+i\sin(\dfrac{\pi/2+2\pi (2)}{3}))

=cos(3π2)+isin(3π2)=\cos(\dfrac{3\pi}{2})+i\sin(\dfrac{3\pi}{2})


Therefore the statement is False.



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