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Complex Conjugate Inequality Proof

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data
    $$
    \left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right |\left ( \left | z \right | +\left | w \right | \right )\leq 2\left | z+w \right |
    $$
    Where z and w are complex numbers not equal to zero.
    2.$$\frac{z}{\left | z \right | ^{2}}=\frac{1}{\bar{z}}$$
    3. The attempt at a solution
    EDIT: INCORRECT WORK

    Alright, the following was my approach. I first tried to split up the problem due to symmetry.
    $$
    \left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right | \left | z \right | \leq \left | z+w \right |
    $$
    $$
    \left | \frac{1}{\bar{z}} + \frac{1}{\bar{w}} \right | \left | z \right | \leq \left | z+w \right |

    $$
    $$
    \left | \frac{\bar{z}+\bar{w}}{\bar{z}\bar{w}} \right | \left | z \right | \leq \left | z+w \right |
    $$
    And I was left with the following
    $$
    \left | \bar{z}+\bar{w} \right | \left | z \right | \leq \left | z+w \right | |
    \bar{z}\bar{w}|
    $$

    I feel like I'm either close, or I've worked the problem entirely the wrong way including conjugates.
     
    Last edited: Sep 21, 2016
  2. jcsd
  3. Sep 21, 2016 #2

    Charles Link

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    I don't think you can split the problem by symmetry in such a manner. I would reconsider that step. If the one term is equal to a where a<1, it means its symmetric pair just needs to be less than 2-a. If you could prove the one term is less than one, it would guarantee the sum is less than 2, but that is probably not the case. Most likely, the single term is not necessarily less than one.
     
  4. Sep 21, 2016 #3
    Hmm, yes I see what you mean. I'm really running out of ideas on how to operate. I guess I could try expanding the binomials, but that looks like a long and arduous task (one I am willing to do).
     
  5. Sep 21, 2016 #4

    Charles Link

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    Suggest you square both sides because |w+z| gives you a square root. Incidentally in your second line in post 1, that should be ## |z|^2 ## in the denominator.
     
  6. Sep 21, 2016 #5
    Yup, looks like I goofed. It does not change anything though does it? Symmetry was no good...
     
  7. Sep 21, 2016 #6

    Charles Link

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    It looks like a complete expansion with squaring is your best approach. I expect terms will cancel on both sides. It is possible you will need to square a second time after that, but I do think this one will not be as difficult as it looks.
     
  8. Sep 21, 2016 #7

    Ray Vickson

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    If you write ##z = r_1 e^{i \theta_1}## and ##w = r_2 e^{i \theta_2}## the left and right-hand sides of your proposed inequality become quite simple functions of ##r_1, r_2, \theta_1, \theta_2##. You can use trigonometric identities and a bit of algebra to get a relatively simple inequality to verify.
     
  9. Sep 21, 2016 #8
    By proposed inequality you mean the initial inequality we are trying to prove, right? The work leading up to the bottom bit is incorrect.
     
  10. Sep 21, 2016 #9

    Ray Vickson

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    Yes, the proposed inequality is the one you are trying to prove.

    When you say the work leading up to the bottom bit is incorrect, whose bottom bit are you referring to? What message number are you citing?

    By using the polar representation I was able to verify your proposed inequality using only elementary algebra, trigonometry and (at one point) elementary calculus. It takes a bit of work, but it really does go through. However, I would not be comfortable with saying much more at this stage, due to PF rules, etc.
     
  11. Sep 23, 2016 #10
    So I went ahead and did the problem yesterday, I believe it ended well. Didn't prove one part though, when I divide both sides by double sines and cosines, I don't check if it's less than 0. https://drive.google.com/open?id=0BxlGAHNyMIneeUpXWEluZWF0TEE
     
  12. Sep 23, 2016 #11
    (Explanations are in spanish, though my guess is you guys will figure it out)
     
  13. Sep 23, 2016 #12

    Charles Link

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    Your ## cos(\theta)cos(\phi)-sin(\theta)sin(\phi) = cos(\theta +\phi) ## so it necessarily has absolute value less than ((edited:) or equal to )1 and won't change the sign of your inequality.
     
    Last edited: Sep 23, 2016
  14. Sep 23, 2016 #13
    True, I think I'll add it to my proof. Thanks Charles (Credit where due)
     
  15. Sep 23, 2016 #14

    Ray Vickson

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    Here is what I did: if ##w = re^{i \lambda}## and ##z = k r e^{i \phi}## (##r,k > 0##) then
    $$L \equiv \left| \frac{w}{|w|} + \frac{z}{|z|} \right| (|w|+|z|) = r(1+k) |e^{i \lambda} + e^{ i \phi}|$$
    and
    $$R \equiv 2 |w+z| = 2 |r e^{i \lambda} + rk e^{i \phi}|$$.
    Without loss of generality we can assume ##\lambda = 0##; if not, just use the fact that
    $$ L = |e^{-i\lambda}|L= r(1+k) |e^{-i \lambda}(e^{i \lambda} + e^{ i \phi})| $$
    and
    $$ R = |e^{- i \lambda}| R = 2 |e^{-i \lambda}(r e^{i \lambda} + rk e^{i \phi})|$$
    use the new variables ##w = r## and ##z = k r e^{i \theta}##, where ##\theta = \phi - \lambda##.

    We have
    $$L = r (1+k) \sqrt{2(1+c)},\; R = 2 r \sqrt{k^2+1+2 k c},$$
    where ##c = \cos(\theta)##. Thus,
    $$ \begin{array}{rcl}R-L &=& r [ 2 \sqrt{1+k^2 + 2kc} - (1+k) \sqrt{2(1+c)}] \\
    &=& r \sqrt{1+c}\, f(k,c),
    \end{array}$$
    where
    $$f(k,c) = 2 \sqrt{ 2k + (k-1)^2/(1+c)} - \sqrt{2} (k+1) $$
    (because ##1+k^2+2kc = (k-1)^2 + 2k(1+c)##).

    For any fixed ##k > 0## the function ##f(k,c)## is strictly decreasing in ##c## on the interval ##-1 < c \leq 1##, so for ##-1 < c < 1## we have ##f(k,c) > f(k,1) = 0##. Thus, ##R-L > 0## for ##-1 \leq c < 1##, and ##R=L## for ##c = 1 \; (\theta = 0)##.
     
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