Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fundamental complex inequality

  1. Jul 11, 2010 #1
    To proove the inequality:

    [tex]\left | \int_a^b f(t) dt \right | \le \int_a^b | f(t) | dt[/tex]

    for complex valued [tex]f[/tex], use the following:

    [tex]\textrm{Re}\left [ e^{-\iota\theta}\int_a^b f(t) dt\right ] = \int_a^b \textrm{Re}[e^{-\iota\theta}f(t)] dt \le \int_a^b | f(t) | dt[/tex]

    and then if we set:

    [tex]\theta = \textrm{arg}\int_a^b f(t) dt[/tex]

    the expression on the left reduces to the absolute value [tex]\square[/tex]


    So the last part, where he says that it reduces to the modulus by using that substitution of theta, I don't get it... how does that reduce to the modulus? :smile:
     
  2. jcsd
  3. Jul 11, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let [tex]z=\int_a^b f(t)dt[/tex]

    Let [tex]\theta=arg(z)[/tex]

    Then we can write [tex]z=Re^{i \theta}[/tex] where [tex]R=|z|[/tex]

    Now look at [tex]e^{-i \theta} z = e^{-i \theta} R e^{i \theta} = R = |z|[/tex]
     
  4. Jul 11, 2010 #3
    Excellent, thanks for the quick reply. :smile:
     
  5. Jul 11, 2010 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    To give a bit more of an exposition, you can think of multiplying a complex number by [tex]e^{i \theta}[/tex] as being the same thing as rotating it in the complex plane by an angle of [tex] \theta[/tex]. In this case what they did was rotate the number so that it lined up with the real axis

    It's pretty clever; I've never seen this proof done like this before
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook