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Complex numbers, proof.

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello

    Assume that we have n complex numbers u: [itex]u_1,u_2,...,u_n[/itex], and n complex numbers v:[itex]v_1,v_2,....v_n[/itex]

    I would like to prove that:
    [itex]|\Sigma_{i=1}^nRe(u_i\bar{v_i})| \le |\Sigma_{i=1}^nu_i\bar{v_i}|[/itex]

    I guess this can be written simpler:
    [itex]|\Sigma_{i=1}^nRe(z_i)| \le |\Sigma_{i=1}^n z_i|[/itex]

    2. Relevant equations



    3. The attempt at a solution
    If n=1. I know that this obviously must hold. But When n is bigger than 1, I am not so sure how to show that it holds. It becomes quit messy, I tried moving the absolute value inside the sum, but id didn't work.
     
  2. jcsd
  3. Mar 24, 2014 #2

    jbunniii

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    I suggest writing each ##z## (you omitted the index by the way) in terms of its real and imaginary components.
     
    Last edited: Mar 24, 2014
  4. Mar 24, 2014 #3

    micromass

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    Hint:

    [tex]\textrm{Re}(z+w) = \textrm{Re}(z) + \textrm{Re}(w)[/tex]
     
  5. Mar 24, 2014 #4

    jbunniii

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    By the way, when working with complex numbers, you should probably use some other letter for your index instead of ##i##, which is already reserved for ##\sqrt{-1}##. Unless you are one of those strange people who use ##j## for that purpose. :smile:
     
  6. Mar 24, 2014 #5
    I got this:

    [itex]|\Sigma z_j|=|\Sigma Re(z_j)+i*Im(z_j)|=|\Sigma Re(z_j)+i*\Sigma Im(z_j)|[/itex]

    Now I would use the triangle inequality if I did not have the i. But I can still say that there is always so that the absolute value of the real part of a complex number must be less than the absolute value of the total number?

    so I get

    [itex]\ge |\Sigma Re(z_j)|[/itex]?
     
  7. Mar 24, 2014 #6

    jbunniii

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    Consider working with the square of the magnitude. ##|a + ib|^2 = ???##
     
  8. Mar 24, 2014 #7
    Do you mean that I should write [itex]c=|\Sigma z_j|[/itex], [itex]a=\Sigma Re(z_j)[/itex], [itex]b=\Sigma Im(z_j)[/itex]
    So then I get [itex]c=|a+bi|[/itex], squaring I get [itex]c^2=|a+ib|^2=a^2+b^2[/itex], then I get:

    [itex]|\Sigma z_j|^2 = (\Sigma Re(z_j))^2+(\Sigma Im(z_j))^2[/itex], and since all are positive we must have that:

    [itex]|\Sigma z_j| \ge |\Sigma Re(z_j)|[/itex]
     
  9. Mar 24, 2014 #8

    jbunniii

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    Yes, the key is that ##a^2 + b^2 \geq a^2##, so taking square roots of each side, we get ##\sqrt{a^2 + b^2} \geq \sqrt{a^2}##. The LHS of this inequality is ##|a + ib|## and the RHS is ##|a|##.
     
  10. Mar 24, 2014 #9
    Thank you for your help!
     
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