Very simple complex question

  • #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.
 

Answers and Replies

  • #2
Dick
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(1/sqrt(2)+i/sqrt(2))^2=1/2+2*i/(sqrt(2)*sqrt(2))-1/2=2*i/2=i.
 
  • #3
HallsofIvy
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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.
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
[tex]-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}[/tex].
 

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