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An algebraic property of complex numbers

  1. Aug 8, 2012 #1
    I'm guessing that if [itex]z\in \mathbb C[/itex], then we have

    \left| z^{-1/2} \right|^2 = |z^{-1}| = |z|^{-1} = \frac{1}{|z|}.

    Proving this seems to be a real headache though. Is there a quick/easy way to do this?
  2. jcsd
  3. Aug 8, 2012 #2
    write z in polar form?
  4. Aug 8, 2012 #3


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    Don't mean to nitpick, but remember that it is for z in ℂ\{0} to start with; some profs.

    may take away points in an exam if you don't specify this.

    But also, remember your square root is not defined everywhere, at least not as a function,

    but as a multifunction, since every complex number has two square roots. I mean, the

    expression z1/2 is ambiguous until you choose a branch.

    Sorry if you already are taking this into account; I am in nitpicking mode, but I

    shouldn't take it out on you :) .
  5. Aug 9, 2012 #4


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