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Exp((i*pi)/4) = ? 
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#1
Sep1009, 11:47 PM

P: 151

Not really homework, but part of a homework problem I am working on.
I know that [tex]e^{i\pi}+1=0[/tex] (Euler's Identity) And also that [tex]e^{i \pi} = e^{i \pi}[/tex] But I'm having trouble understanding [tex]e^{\frac{i \pi}{4}}[/tex] In the complex plane this is a clockwise rotation around the origin of [tex]\frac{\pi}{2}[/tex] radians. But I think it should reduce to some real constant which I am having trouble finding. In Mathematica, I get two different answers... [tex]N[e^{\frac{i \pi}{4}}] = 0.707107  0.707107 i[/tex] This implies that [tex]e^{\frac{i \pi}{4}} = \frac{1}{\sqrt{2}}(1i)[/tex] which seems wrong to me. The other answer given is: Simplify[tex][e^{\frac{i \pi}{4}}] = (1)^{\frac{3}{4}}[/tex] But this reduces to 1, which I believe is probably the correct answer. Is the first result just spurious rounding? Can I just write the following identity as true? [tex]e^{\frac{i \pi}{4}} = 1[/tex] 


#2
Sep1009, 11:59 PM

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P: 25,235

e^(i*pi/4) isn't 1 by a long shot. e^(i*x) is cos(x)+i*sin(x). e^(i*pi/4)=cos(pi/4)+i*sin(pi/4). Both of your Mathematica answers are correct. In fact they are the same (after rounding). And (1)^(3/4) is NOT 1. Or 1 either.



#3
Sep1109, 12:00 AM

P: 1,104

Eh the first answer looks correct to me, since .707 is around sqrt(2)/2. You do know the Euler's formula, which is where those identities are usually derived from right?



#4
Sep1109, 12:05 AM

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P: 21,397

Exp((i*pi)/4) = ?



#5
Sep1109, 12:22 AM

P: 151

Thanks Mark44 and all. This makes a lot more sense to me now.



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