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Roots of unity, Roots of complex equations of the form z^n = 1

  1. Nov 18, 2014 #1
    Dear all,


    please see the page above, (Alan F, Beardon, Abstract Algebra and Geometry). On the page, Theorem 3.5.2 says that the set of Complex numbers from ## z^n = 1 ##, where ## |z| = 1 ## forms a group w.r.t multiplication. I want to know if.....

    The inverse of all elements are their complex conjugates?

    inverse of the identity, z = 1∠0 is its own inverse??

    is an inverse missing when n is even??
    In the above diagram, n = 8. z = 1 is the inverse of itself. the three complex points above the real axis have inverses as the three complex points below the real axis. What about z = -1?? where is its inverse?? is it z = 1? why???
  2. jcsd
  3. Nov 18, 2014 #2


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    well when you multiply two complex numbers their angles add and their lengths multiply, so does that answer your question about what happens when you multiply two complex conjugates of length one?
  4. Nov 23, 2014 #3
    The inverse of z = -1, the point on the Leftmost side of the circle above is its own inverse. It's because in rectangular form its ## z = -1 +0i ## so its conjugate is ##z' = -1 - 0i## which lies on the same point as z.
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