Math Help - 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:

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

rearrange this to:

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

so:

$z=2-3 u$

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

$
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:

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

so:

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

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

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

CB