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Complex conjugate proof, i think

  1. Jan 17, 2015 #1
    1. The problem statement, all variables and given/known data

    prove that sqrt2|z| greater than or equal to |Rez| + |Imz|

    2. Relevant equations
    |z|^2 = x^2 + y^2
    Rez=x, Imz=y


    3. The attempt at a solution
    so far I've worked it down to this.
    2(x^2 + y^2) greater than or equal to x^2 + 2xy + y^2

    I've used a few different values for x and y and it holds true but i don't know if that right or that's how it should be written.
    any input would be appreciated.
     
    Last edited: Jan 17, 2015
  2. jcsd
  3. Jan 17, 2015 #2

    mfb

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    Where does that come from? It is a trivial statement right now, but I think there is a mistake. Fix it, and prove the relationship with the usual rules for manipulating inequalities.
     
  4. Jan 17, 2015 #3
    i actually just fixed it.
     
  5. Jan 17, 2015 #4

    mfb

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    Good. You can write it as inequality and reduce it to something you know to be true.
     
  6. Jan 17, 2015 #5
    thats where im stuck. i keep trying to break it down but i keep ending up with a jumbled mess

    x greater than or equal to x/2 +y - y^2/2x ?
     
  7. Jan 17, 2015 #6
    ok i dont know if i can do this but here it is

    x^2-2xy +y^2 graeter than or equal to 0

    (x-y)^2 = 0

    sqrt each side, move some stuff and...

    x greater than or equal to y.

    and thats true right?
     
    Last edited: Jan 17, 2015
  8. Jan 17, 2015 #7
    so...
    (x-y)(x-y) greater than or equal to 0
    divide one of them out
    x-y greater than or equal to zero
    x greater than or equal to y

    i dont understand how that is intuitively true though.
     
    Last edited: Jan 17, 2015
  9. Jan 17, 2015 #8

    Dick

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    You can't divide one of the them out without knowing the sign of x-y. And I think you might be missing a step in going from sqrt(2|z|)>=|Re(z)|+|Im(z)| to x^2-2xy-y^2>=0. Pay attention to the absolute values on Re(z) and Im(z).
     
    Last edited: Jan 17, 2015
  10. Jan 17, 2015 #9
    really? ((x-y)(x-y))/(x-y) doesnt equal x-y ?
     
  11. Jan 17, 2015 #10

    Dick

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    Yes, it does but a*a>=0 doesn't imply that a>=0.
     
  12. Jan 17, 2015 #11
    well then... i have no idea what to do with this problem anymore.

    first day of class..... completely lost.
     
  13. Jan 17, 2015 #12

    Dick

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    (-1)(-1)>=0. But (-1)>=0 is false. If you divide both sides by a negative you have to reverse the inequality. You got to the correct point of wanting to prove that x^2-2xy-y^2>=0 (though you may have skipped a step, can you spell them all out). You can factor that into (x-y)^2>=0 which shows what you want to show. You don't have to (and can't!) show x>=y.
     
  14. Jan 17, 2015 #13
    so what youre saying is (x-y)^2 >= 0 is always true. which i can see. because (x-y)^2 is always positive. ahhhh thank you.

    problem 1 of 10 down haha

    the last time i did proofs was in precalc 2 and those were trig proof... about 3 years ago and 5 math classes ago haha so i feel new to this stuff. its actually for a class titled "intermediate mathematical methods", PHYS395. i just havent seen this stuff in a good while.

    thank you again.
     
  15. Jan 17, 2015 #14

    Dick

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    Ok, but what I've been trying to say is that your original statement after squaring leads to 2(x^2+y^2)>=x^2+2|x||y|+y^2. Do you see why? There's an extra bit to turn that into x^2-2xy-y^2>=0.
     
  16. Jan 17, 2015 #15
    yeah i see why. on my paper i have it written out step by step, including that one.

    i guess i was a bit unclear on when to stop the algebra haha but i see now where we stopped is the logical place to stop.
     
  17. Jan 17, 2015 #16

    Dick

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    Ok, I'll trust you.
     
  18. Jan 17, 2015 #17
    i can take a picture of my notes if you like. its some printing paper covered in mess though haha but its in there.

    next time ill just upload a copy of my work to photobucket and post it into the "attempt at a solution" area... im sure you guys will see more of me.
    my first year hitting 300 level physics courses and a 400 level math course.
    thank you again.
     
  19. Jan 17, 2015 #18

    Dick

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    Yes, post all of the details of your solution. Always best. See you later!
     
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