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Math Help - cube root of unity..compex number

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Shivanand View Post
    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:

     <br />
z=-1,\ 2-2w,\ 2-2w^2<br />

    CB
    Last edited by CaptainBlack; December 16th 2008 at 03:50 AM.
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by Shivanand View Post
    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
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