1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Very simple complex question

  1. Oct 2, 2007 #1
    According to this equation [tex] \exp^{iy} = \cos y + \iota \sin y \mbox{ => } \exp^{\frac{\pi \iota}{2}} = \iota \ \mbox{,} \ \\ \exp^{\pi \iota} = -1 \ \ \exp^{\frac{-\pi \iota}{2}} = -\iota \ \ \exp^{2 \pi \iota} = 1 \\ [/tex]. What then is [tex] \sqrt{\iota} = \iota^{\frac{1}{2}} \mbox{ equal to ?} \\ [/tex]
    Does not [tex] \exp^{\frac{\pi \iota}{4}}=\cos\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} +\frac{\iota}{\sqrt{2}} \ \mbox{ which is not =} \ \sqrt{\iota} \\[/tex]
    I find it hard to see that it equals [tex] \exp^{\frac{\pi \iota}{4}}\\[/tex]. Help would be welcome. Thanks.
  2. jcsd
  3. Oct 2, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

  4. Oct 2, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Why do you think [itex]\sqrt{2}/2 +i\sqrt{2}/2[/itex] is not equal to [itex]\sqrt{i}[/itex]. Dick did the calculations but it's just a little bit hard to read. In Tex:
    [tex]\left(\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\right)[/tex]
    [tex]= \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}\frac{i\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\frac{i\sqrt{2}}{2}[/tex]
    [tex]= \frac{1}{2}+ \frac{i}{2}+ \frac{i}{2}- \frac{1}{2}= i[/tex]
    Of course, like any non-zero number, i has two square roots. The other is
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Very simple complex question