I can't see the next step:
$D^\alpha e^{ix} = i^{\alpha}e^{ix} = e^{i\alpha \frac \pi2}e^{ix}$
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Did you forget to put the $e^{ix}$ at the end? Is this question currently incomplete? Please give some context: what is supposed to be shown here and what don't you understand? – Henry T. Horton May 14 '13 at 01:36
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Yes, thanks @HenryT.Horton – David May 14 '13 at 01:37
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1$i=e^\frac{i\pi}{2}$, is that your question? – Alex R. May 14 '13 at 01:54
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Yes @Alex. I don't understand this step. – David May 14 '13 at 03:15
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1Then this will be helpful: http://en.m.wikipedia.org/wiki/Euler%27s_formula – Alex R. May 14 '13 at 03:26
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Could you specify which definition of fractional derivative you are using? The are several. Depending the one you are using $D^{\alpha}e^{ix}=i^{\alpha}e^{ix}$ is wrong. – Ambesh May 19 '13 at 07:45
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1See here. You should know what definition you are using for fractional derivative. Check the link. – Mhenni Benghorbal May 30 '13 at 00:04
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A related problem. – Mhenni Benghorbal May 30 '13 at 00:20
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Using Euler's identity $$e^{ix} = \cos x + i \sin x$$ A complex number must be in the form: $z = x+iy$, then $e^z = e^{x+iy} = e^x+e^{iy} = e^x(\cos y + i \sin y)$ and I want that this last equality be $i$, then, $e^x(\cos y + i \sin y) = i$ only if $x = 0$ and $y = \frac \pi2$.
David
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