1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]

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

    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}}


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook