Answer to Question #137496 in Algebra for Sarita bartwal

Question

There is one and only one complex cube root of 1.
True or false with correct explanation

Answer 1

False.

There are three complex cube roots of 1.

$1>0$ so we can find these cube roots:

$x_1=\sqrt{1}=1$

$x_2=\sqrt{1}*(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i)=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$

$x_3=\sqrt{1}*(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i)=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$

where

$\sqrt{1}=1$ is a simple cube root of 1.

$(\sqrt{1})^3=1$

$(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i)^2=\dfrac{\sqrt{3}}{2}i-\dfrac{2}{4}$

$(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i)*(\dfrac{\sqrt{3}}{2}i-\dfrac{2}{4})=1$