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Homework Help: Exp((-i*pi)/4) = ?

  1. Sep 10, 2009 #1
    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}}(1-i)[/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. jcsd
  3. Sep 10, 2009 #2


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    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.
  4. Sep 11, 2009 #3
    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?
  5. Sep 11, 2009 #4


    Staff: Mentor

    No, the rotation is by [tex]\frac{\pi}{4}[/tex] radians. Was the 2 in the denominator a typo?
    Why would you think that? Just because [tex]e^{-i \pi}[/tex] is a real constant, doesn't mean that the other one is also a real constant.
    These are two representations of the same complex number. The first is an approximation and the second is exact.
    I am not familiar with Mathematica, so I don't know the difference between the N command and the Simplify command.
    No it doesn't, and 1 is not the correct answer. Let's look at -(-1)3/4 a little more closely. Before the final sign change, you have (-1) to the 3/4 power. That is the same as -1 cubed (still -1), which we take the 4th root of. This is not a real number, since there is no real number that when squared, and then squared again, yields a negative number. The final step is to change the sign of this (nonreal) number, which still doesn't give us 1, as you claimed.
    Absolutely not.
  6. Sep 11, 2009 #5
    Thanks Mark44 and all. This makes a lot more sense to me now.
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