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Homework Help: Solving (e[SUP]i[/SUP])[SUP]i[/SUP]

  1. Aug 21, 2011 #1
    Hey guys,

    I have the following question that I am trying to solve. Would really like to solve it on my own first so I'd appriciate hints first! :D

    The problem statement, all variables and given/known data

    Solve (ei)i

    The attempt at a solution

    (ei)i = (cos pi/pi + isin pi/pi)i
    = (cos i + isin i) (Euler's Formula)

    I'm not sure if I'm even right till this point but this is the furthest that I have managed to come so far. ):

    Thanks in advance guys!
  2. jcsd
  3. Aug 21, 2011 #2


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    Science Advisor

    It might be more helpful to note that if [itex]y= (e^i)^i[/itex] then [itex]ln(y)= i ln(e^i)= i^2= -1[/itex]. And, of course, [itex]ln(re^{i\theta}= ln(r)+ i(\theta+ 2k\pi)[/itex].
  4. Aug 21, 2011 #3
    Wouldn't that leave me with ln y = -1 and ln y = ln i?

    Anyway thanks for your quick reply HallsofIvy! Your quick reply is very much appriciated as usual! :D
  5. Aug 21, 2011 #4


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    Staff: Mentor

    (ei)i = ei*i = e-1 = ......
  6. Aug 21, 2011 #5
    I thought of this but didn't think it would be this simple hahaha. So just the calculator value of e-1? :D
  7. Aug 21, 2011 #6


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    No, "[itex]2k\pi[/itex] is always a multiple of [itex]2\pi[/itex], not [itex]\pi. You would only get ln y= -1.

    Ouch! Of course!
  8. Aug 21, 2011 #7


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    Homework Helper

    I had a post here which I deleted, as I missed the same thing. HallsofIvy's point is correct in general, in that complex numbers can have an infinite number of complex logarithms (think of what happens with taking roots with DeMoivre's Theorem, where there are six sixth roots of 1 , and so forth; it gets worse with logarithms).

    However, they were cute with this problem because all of the logarithms fall in the same spot on an Argand diagram.

    If we work with the complex exponential forms for sine and cosine, e.g., [itex]\cos z = \frac{e^{iz} + e^{-iz}}{2} [/itex], we just get one real value for ( cos i + i sin i ) . Other staged exponentials involving complex numbers may not resolve so neatly...
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