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Why is this false?

  1. Sep 8, 2014 #1
    (e^{1+i*2*pi})^{1+i*2*pi}=e^{(1+i*2*pi)(1+i*2*pi)}
    why is this false (according to Wolfram Alpha) ?
     
  2. jcsd
  3. Sep 8, 2014 #2
    Because the exponent laws are only true for real numbers.
     
  4. Sep 8, 2014 #3
    ##\begin{align}
    (e^{1+i2\pi})^{1+i2\pi}&=(e\cdot e^{i2\pi})\cdot(e\cdot e^{i2\pi})^{i2\pi}\\
    &=(e\cdot1)(e\cdot1)^{i2\pi}=e
    \end{align}##

    but

    ##\begin{align}
    e^{(1+i2\pi)(1+i2\pi)}&=e^{1-4\pi^2+i4\pi}\\
    &=e^{1-4\pi^2}\cdot1\neq e
    \end{align}##

    The complex exponential is defined (I recall) so that ##e^{z_1+z_2}=e^{z_1}e^{z_2}##. Beyond that the algebra needs to be derived. This exponent "law" is only a law for real numbers.
     
  5. Sep 8, 2014 #4

    WWGD

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    Science Advisor
    Gold Member

    Because you have to choose a branch of log to define ## z^a ## when z is complex with non-zero imaginary part and a is not a rational number. Take, e.g.,

    (eiπ/4))iπ/4 .

    Let z =eiπ/4 . Then

    (eiπ/4))iπ/4=ziπ/4:=ezlog(iπ/4) . Using the

    ##Logz## branch, i.e., the main branch (because the argument of iπ/4 is precisely π/4 in the main branch Logz), we have

    Log(iπ/4)=ln(1)+iπ/4=iπ/4 and then it works out:

    (eiπ/4)iπ/4=(cos(π/4)+isin(π/4))iπ/4= ##( \sqrt 2/2+isin \sqrt 2/2)^{i\pi/4}= ( \sqrt 2/2+isin \sqrt 2/2)^{ln(1)+i\pi/4} =e^{-\pi^2/16}##

    . But if you want to use ##2 \pi ## as argument, then you need to work in a branch where ## 2\pi ## makes sense.
     
    Last edited: Sep 8, 2014
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