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Absolute value of exponetial?

  1. Mar 16, 2013 #1
    I want to calculate [itex]|e^{a^{2} + \frac{it}{m\hbar}}|^{2}[/itex]


    i is imaginary unit.


    my trie:

    [itex]a^{2} + \frac{it}{2m\hbar}[/itex] is a complex number so its module is:


    [itex]\sqrt{a^{4} + \frac{t^{2}}{m^{2}\hbar^{2}}}[/itex]

    = [itex]\sqrt{a^{4}(1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}})}[/itex]

    [itex]a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}[/itex]


    So the solution is:


    [itex](e^{a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}})^{2}[/itex]

    =
    [itex]e^{2a^2\sqrt{1 + \frac{t^{2}}{m^{2}\hbar^{2}a^{4}}}}[/itex]




    My friend said that is is definitely wrong. And i actually think that it is wrong.

    can somebody tell me where?
     
  2. jcsd
  3. Mar 16, 2013 #2

    Fredrik

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    Gold Member

    Is there a theorem that says that ##|e^z|=e^{|z|}## for all complex numbers z? Since I'm asking, you can guess that the answer is no. I suggest that you verify this by finding a counterexample.

    Do you know anything about the exponential function that would allow you to rewrite ##e^{a^2+\frac{it}{m\hbar}}## in a different way?
     
  4. Mar 16, 2013 #3
    yup

    [itex]e^{(x+b)} =e^{x}.e^{b}[/itex]

    so i am going to have a e^(real part)e^(imaginary part)

    absolute value of e^(imaginary) = 1 right (euler's formula)

    so the solution is [itex]|e^{(real part)}|^2 = e^{(2(a^{2}))}[/itex]
     
    Last edited: Mar 16, 2013
  5. Mar 16, 2013 #4

    Dick

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    Yes, correct.
     
  6. Mar 16, 2013 #5
    ok thanks for help
     
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