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Math Help - Find all complex numbers satisfying z^6=1

  1. #1
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    Find all complex numbers satisfying z^6=1

    Hi guys.

    "Find all complex numbers satisfying z^6=1"

    The hint is to factorise z^6-1 as a difference of squares.

    (z^3+1)(z^3-1)
    z(z^2+\frac{1}{z}) \cdot z(z^2-\frac{1}{z})

    This seems incorrect.

    The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.

    I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the cis\Theta where \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi.

    I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

    I've asked several people about this problem, but they can't seem to distinguish copying from understanding.

    I would love to really understand the solution to this problem.

    Can you help?
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  2. #2
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    Quote Originally Posted by lucius View Post
    Hi guys.

    "Find all complex numbers satisfying z^6=1"

    The hint is to factorise z^6-1 as a difference of squares.

    (z^3+1)(z^3-1)
    z(z^2+\frac{1}{z}) \cdot z(z^2-\frac{1}{z})

    This seems incorrect.

    The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.

    I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the cis\Theta where \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi.

    I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

    I've asked several people about this problem, but they can't seem to distinguish copying from understanding.

    I would love to really understand the solution to this problem.

    Can you help?
    I'd go with the sum/difference of two cubes afterwards.

    z^6-1=0

    (z^3-1)(z^3+1)

    z^3-1 = (z-1)(z^2+z+1)

    z^3+1 = (z+1)(z^2-z+1)

    z^6-1 = (z-1)(z+1)(z^2+z+1)(z^2-z+1)

    Use the quadratic formula to solve the last two terms which should give 6 solutions
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  3. #3
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    Quote Originally Posted by lucius View Post
    "Find all complex numbers satisfying z^6=1"
    I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the cis\Theta where \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi.

    I'm very interested in understanding the solution with De Moivre's theorem.
    Define \xi  = e^{\frac{{i\pi }}{3}}  = cis\left( {\frac{\pi }{3}} \right).
    Now the six roots are \xi ^k ,\;k = 0,1,2, \cdots ,5.
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  4. #4
    kjf
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    math1051 assignment 1 @ UQ?
    I'm stuck, too. I think i'm way out of my element here, but I just plugged in some values into de moivre's theorem.

    I don't fully understand complex numbers, and I too am keen to know HOW to solve a question like this, rather than have someone say "here's the answer, off you go". Will be watching this thread.
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  5. #5
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    Isn't the answer provided by e^{i\pi} clear enough?

    It's simple application of factorisation rules DOTS and Sum/Difference of two cubes, and null factor law...

    It's set equal to 0, so that means each factor could be 0. Solve for z.
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    Quote Originally Posted by lucius View Post
    I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.
    First, we express the RHS in the form | z|(\cos\theta+i\sin\theta), where |z| is of course the modulus, and \theta the argument. Since |1| = 1, and \arg{1} = 0, we have:

    1 = \cos(0)+i\sin(0) \Rightarrow z^6 = \cos(0)+i\sin(0) \Leftrightarrow  z = \left\{cos(0)+i\sin(0)\right\}^{\frac{1}{6}}

    De Moivre's theorem states that: \left\{\cos(\phi)+i\sin(\phi)\right\}^n = \cos{n}\phi+i\sin\phi

    Or more relevantly: \left\{\cos(\phi)+i\sin(\phi)\right\}^{\frac{1}{n}  } = \cos\left[\dfrac{\phi+(n-1)2\pi}{n}\right]+i\sin\left[\dfrac{\phi+(n-1)2\pi}{n}\right]

    So \left\{\cos(0)+i\sin(0)\right\}^{\frac{1}{n}} = \cos\left[\dfrac{(n-1)2\pi}{6}\right]+i\sin\left[\dfrac{(n-1)2\pi}{6}\right]

    We just plug n = 0, 1, 2,...5 to get the six values and then convert them into the Cartesian form, i.e. \cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\p  i}{3}\right) = \frac{1}{2}+\frac{\sqrt{3}}{2}i\.

    Note that all the subsequent values after n = 5 are repetitions, i.e n = 0 has the same value as n = 6.
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    Quote Originally Posted by kjf View Post
    math1051 assignment 1 @ UQ?
    I'm stuck, too. I think i'm way out of my element here, but I just plugged in some values into de moivre's theorem.
    Haha yeah math1051. I did the same thing that was in the first resposnse, from e^(i*pi) - difference of 2 squares then differnces of two cubes. Its way easier that looking at D.M's theorem - way faster! and you still get the 6 solutions, so really, whats not to love? haha
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    DeMoivre's Theorem isn't too bad as long as you remember that the roots are all evenly spaced around a circle.


    So, in this case, remember that z^6 = 1 = \cos{2\pi} + i\sin{2\pi}.


    Therefore:

    z = \left[1\left(\cos{2\pi} + i\sin{2\pi}\right)\right]^{\frac{1}{6}}

     = 1^{\frac{1}{6}}\left(\cos{\frac{2\pi}{6}} + i\sin{\frac{2\pi}{6}}\right)

     = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}.


    Since they are evenly spaced around the circle and there are 6 roots, that means they are all separated by an angle of \frac{2\pi}{6} = \frac{\pi}{3}.


    So the other roots are:

    \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}}

    \cos{\frac{3\pi}{3}} + i\sin{\frac{3\pi}{3}}

    \cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}

    \cos{\frac{5\pi}{3}} + i\sin{\frac{4\pi}{3}}

    and

    \cos{\frac{6\pi}{3}} + i\sin{\frac{6\pi}{3}}.


    If you convert each of these to cartesians, you should get the same answers as those found by factorising and using Null Factor Law.
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  9. #9
    kjf
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    I've now done it both ways (ie; diff of squars and DM theorum).
    Both worked.

    to anyone else doing the 1051 assignment, what in the hell is he asking for in question 10?

    "Suppose y = cos(x), where 0<= x <= pi/2, obtain sin(x) as a function of y for the same interval.

    Is he just asking, what sin function will give us the same "plot" as the cos function? something like 1-sin(x), i've done it, i just can't find it. But I don;t know if that's what we have to do?
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  10. #10
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    Quote Originally Posted by kjf View Post
    I've now done it both ways (ie; diff of squars and DM theorum).
    Both worked.

    to anyone else doing the 1051 assignment, what in the hell is he asking for in question 10?

    "Suppose y = cos(x), where 0<= x <= pi/2, obtain sin(x) as a function of y for the same interval.

    Is he just asking, what sin function will give us the same "plot" as the cos function? something like 1-sin(x), i've done it, i just can't find it. But I don;t know if that's what we have to do?
    You should know that

    \sin^2{x} + \cos^2{x} = 1

    so

    \sin^2{x} = 1 - \cos^2{x}

    \sin{x} = \pm \sqrt{1 - \cos^2{x}}.


    Since you are only dealing with the first quadrant, you know that sine is positive, so

    \sin{x} = \sqrt{1 - \cos^2{x}}

     = \sqrt{1 - y^2}.
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  11. #11
    kjf
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    Thankyou so much Prove It. I'm kind of a rope-learner, for things I have never fully encountered before I need to be shown how to do it. In fact, all I had to do was look at your post, remember than sin^2x + cos^2x = 1 and I had a go. compared my answer to yours and it came out the same!

    Thanks again!
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