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Finding the cube root of 1 using Euler's formula

  1. Aug 11, 2012 #1

    lo2

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    1. The problem statement, all variables and given/known data

    I have found this video where there is this problem:

    Find the cube root of 1

    2. Relevant equations

    I have found this video:


    Where from 8:02 to the end she solves this problem.


    3. The attempt at a solution

    My question is why is she making such a big fuss about this? Is the cube root of 1 not just simply only 1? And if not could someone please explain what she is trying to explain?
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Aug 11, 2012 #2

    Mentallic

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    Homework Helper

    so we're looking for the value of z where [itex]z=1^{1/3}[/itex]
    if we cube both sides of this equality, we get [itex]z^3=1[/itex], which is equivalent to [itex]z^3-1=0[/itex]
    Can you factorize that expression?
     
  4. Aug 11, 2012 #3
    It is not simply only 1 because it is not simply only 1. There are three different complex cubic roots of 1, and that's what is shown in the video.
     
  5. Aug 11, 2012 #4

    AGNuke

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    Gold Member

    You are true. Each polynomial function have a degree, and equal number of roots, whether real or imaginary. Cubic expressions always have 3 roots. So, there are 3 cube roots of 1. They are 1, ω, ω2. It can be found by simple quadratic as follows. I hope you know the identity of factorizing (a-b)3
    [tex]x^3-1=0\Rightarrow (x-1)(x^2+x+1)=0[/tex]

    Solutions of the quadratic polynomial factor.
    [tex]x=\omega = \frac{-1-i\sqrt{3}}{2};x=\omega ^2=\frac{-1+i\sqrt{3}}{2}[/tex]

    This can also be done by Euler's formula,
    [tex]z^3=e^{i2n\pi}\Rightarrow z=e^{\frac{i2n\pi}{3}}[/tex]

    By putting n = 0, 1, 2 (3 will yield same result as 0, 4 as 1...)

    [tex]z_1=e^{\frac{i0\pi}{3}};z_2=e^{\frac{i2\pi}{3}};z_3=e^{\frac{i4\pi}{3}}[/tex]

    By the identity, [itex]e^{i\theta}=cos\theta+isin\theta[/itex], you get z1, z2 and z3
     
    Last edited: Aug 11, 2012
  6. Aug 11, 2012 #5

    lo2

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    Ok. I can follow you on the first way of finding the roots of the cubic root!

    I am not sure I can follow why these two things are equal to each other could you please elaborate?
     
  7. Aug 11, 2012 #6

    eumyang

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    AGNuke took the cube root of both sides, or raised both sides to the 1/3 power. However, there is an i missing in the result, so the equation after the right arrow should be:
    [tex]z = e^{i2n\pi/3}[/tex]
    This is not quite right, because the middle root written above is a cube root of -1, not 1. The three roots should be
    [itex]z_1 = e^{i \cdot 0\pi/3}[/itex]
    [itex]z_2 = e^{i \cdot 2\pi/3}[/itex]
    [itex]z_3 = e^{i \cdot 4\pi/3}[/itex]
     
    Last edited: Aug 11, 2012
  8. Aug 11, 2012 #7

    AGNuke

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    Gold Member

    Corrected!
    What you can't follow, that [itex]e^{i\theta}=\cos\theta+i\sin\theta[/itex]?
     
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