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Homework Help: Simple equality help e^(-ix)

  1. Sep 20, 2015 #1


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    1. The problem statement, all variables and given/known data
    Very simple problem I'm having:

    The problem is: Show 1/e^(ix) = e^(-ix)

    2. Relevant equations
    * is the complex conjugate.
    e^(ix) = cos(x)+isin(x)

    3. The attempt at a solution
    RHS = [itex] e^{-ix} = e^{ix^*} = (cos(x)+isin(x))^* = cos(x)-isin(x)[/itex]

    LHS = [itex] \frac{1}{e^{ix}} = \frac {1}{cos(x)+isin(x)} = cos(x)-isin(x)[/itex]

    My question:
    1) How does the LHS go from 1/(cos(x)+isin(x)) to cos(x)-isin(x)? I assume there is some trig property involved here?
    2) Is it safe to state that e^(-ix)=e^(ix)* where * is the complex conjugate? Do I need to really prove anything here?
    Last edited by a moderator: Sep 21, 2015
  2. jcsd
  3. Sep 20, 2015 #2

    Ray Vickson

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    For real ##a## and ##b##, do you know how to express ##1/(a + ib)## as ##A + iB##, where ##A,B## are also real?
  4. Sep 20, 2015 #3


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    There it is. The connection.

    We multiply numerator and denominator by the complex conjugate to see
    cos(x)-isin(x) over cos^2(x)+sin^2(x) which we know by trig is just the numerator.

    That iss Q1 cleared.
  5. Sep 20, 2015 #4
    1. That's because $$\frac{1}{\cos x+i\sin x}=\frac{\cos x-i\sin x}{(\cos x +i\sin x)(\cos x-i\sin x)}=\ldots$$ Then use the Pythagorean identity.
    2. I would say: it isn't safe.
  6. Sep 20, 2015 #5
    $$ e^{-ix} = e^{(ix)(-1)} = (e^{ix})^{-1} = \frac 1{e^{ix}} $$
  7. Sep 20, 2015 #6


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    Ah, Q2) isn't too hard to prove as well.

    e^(-ix) = cos(-x)+isin(-x) = cos(x)-isin(x).

    This is complete now.

    Kind regards for your words of motivation.
  8. Sep 21, 2015 #7


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    I don't see any reason to change to the trig form [itex]\left(e^{-ix}\right)\left(e^{ix}\right)= e^{I(x- x)}= e^0= 1[/itex] is sufficient to prove that [itex]e^{ix}[/itex] is the multiplicative inverse of [itex]e^{ix}[/itex].
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