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Set of all values of sqrt(i)

  1. Feb 15, 2013 #1
    Why is it that set of all values of ##log(√i)## is ##(\frac{1}{4} + n)\pi i## and not ##(\frac{1}{4} + 2n)\pi i## since ##√i = e^\frac{ \pi i}{4}##

    and ##log(z) = ln|z|+ i(Arg(z) + 2 n \pi)##
     
  2. jcsd
  3. Feb 16, 2013 #2
    Looking at it, I get:
    [tex]
    \log \left[\sqrt{i}\right] = \log\left\{\exp\left[\frac{i}{2}\left(\frac{\pi}{2}+2\pi n\right)\right]\right\} = \frac{i}{2}\left(\frac{\pi}{2} + 2\pi n\right)\log e = i\left(\frac{\pi}{4} + \pi n\right)
    [/tex]
    where [itex] n[/itex] is an integer. It looks like you might have used a period of [itex]\pi[/itex] rather than [itex]2\pi[/itex].
     
  4. Feb 16, 2013 #3

    jbunniii

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    If we exponentiate ##(\frac{1}{4} + n)\pi i## and then square it, we get
    $$\left(\exp\left(\left(\frac{1}{4} + n\right)\pi i\right)\right)^2 =
    \left(\exp\left(\frac{i\pi}{4}\right)\exp\left(in \pi \right)\right)^2 =
    \exp\left(\frac{i\pi}{2}\right)\exp\left(i2n \pi\right) = i$$
    so every number of the form ##(\frac{1}{4} + n)\pi i## is a solution. Your proposed answer misses all of the solutions for which ##n## is odd.
     
  5. Feb 16, 2013 #4

    haruspex

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    True, but if √ is taken to denote the principal square root (which would be standard, yes?) then only the 2n solutions arise.
     
  6. Feb 16, 2013 #5

    jbunniii

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    Yes, if the principal square root is intended. The "set of all values of sqrt(i)" in the subject led me to presume otherwise, but that may not have been warranted. Bachelier, what do you intend √ to mean in this context?
     
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