# Thread: cube root of unity..compex number

1. ## cube root of unity..compex number

this question asked so many time in final year exam ,, but we were not able to solve this...... If 1,w,w^2 are the cube roots of unity, then find the roots of (z – 2)3 + 27 = 0.

2. Originally Posted by Shivanand
this question asked so many time in final year exam ,, but we were not able to solve this...... If 1,w,w^2 are the cube roots of unity, then find the roots of (z – 2)3 + 27 = 0.
If:

$\displaystyle (z-2)^3+27=0$

rearrange this to:

$\displaystyle (z-2)^3=-27$

so:

$\displaystyle z=2-3 u$

where $\displaystyle u$ is a cube root of unity is a root, so the required roots are:

$\displaystyle z=-1,\ 2-2w,\ 2-2w^2$

CB

3. ## cube root of unity

If 1,w,w^2 are the cube roots of unity, then find the roots of (z – 2)^3 + 27 = 0. Plz ellaborate ur answer ,i am unable to understand . and also tell me from where can i find information for this.

4. Originally Posted by Shivanand
If 1,w,w^2 are the cube roots of unity, then find the roots of (z – 2)^3 + 27 = 0. Plz ellaborate ur answer ,i am unable to understand . and also tell me from where can i find information for this.
If:

$\displaystyle (z-2)^3=-27=27 e^{(\pi+2n\pi)i},\ n=0, \pm1,\ \pm 2,\ ..$

so:

$\displaystyle (z-2)=3 e^{(\pi+2n\pi)i/3},\ n=0, \pm1,\ \pm 2,\ ..$

but only three of the exponential terms on the right are distinct and these are the cube roots of $\displaystyle -1$, and the three distinct values are $\displaystyle -1$, $\displaystyle -w$, $\displaystyle -w^2$, from which the previous result follows.

This uses Eulers formula for complex numbers and the laws of powers.

CB