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Trouble with Complex Exponentials

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if Im(z) ≥ 0, then |[itex]{{\rm e}^{i \left( x+iy \right) }}[/itex]| ≤ 1.

    Let R > 1 be a real constant

    Now deduce that [itex]{\frac {{{\rm e}^{-{\it Im} \left( x+iy \right) }}}{ \left| \left( x+
    iy \right) ^{4}+1 \right| }}
    [/itex] ≤ [itex] \left( {R}^{4}-1 \right) ^{-1}[/itex] for z on the semi-circle {z [itex]\in[/itex] ℂ: |z| = R, Im(z) ≥ 0}



    3. The attempt at a solution

    The imaginary part of z ≥ 0 is just y ≥ 0.

    Multiplying the exponential through, you get, [itex]{{\rm e}^{ix-y}}[/itex].

    I realise that taking the modulus of this will make the e^ix disappear, leaving only e^-y, which clearly shows then as y ≥ 0:

    e^-y ≤ 1

    I'm not too sure how to get the modulus of that exponential.

    N.B: The modulus applies to the entire fraction, I cannot seem to get the modulus including the entire fraction for some reason, it's just appearing on the denominator
     
    Last edited: Apr 22, 2012
  2. jcsd
  3. Apr 22, 2012 #2
    I tried a few methods, with this I got what looked correct, but I wasn't really sure of what I was doing.

    [itex]{{\rm e}^{ix}}{{\rm e}^{-y}}[/itex]

    [itex]
    {{\rm e}^{-y}} \left( \cos \left( x \right) +i\sin \left( x \right)
    \right)
    [/itex]

    Taking the absolute value of that:

    Square Root of [itex][\left( {{\rm e}^{-y}} \right) ^{2} \left( \cos \left( x \right)
    \right) ^{2}+ \left( {{\rm e}^{-y}} \right) ^{2} \left( \sin \left( x
    \right) \right) ^{2}][/itex]

    Square Root of [itex][{{\rm e}^{-2\,y}}*(1)][/itex]

    Which is [itex] {{\rm e}^{-y}}

    [/itex]

    Voila? So now we can see that as y increases the value of e^-y is going to decrease, with its maximum value when y = 0 as per Im (z).

    Edit: Forgot to add the second part of the question earlier, just added it in, not sure on how to proceed with that part.
     
    Last edited: Apr 22, 2012
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