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Ug, a couple more tough problems

  1. Feb 10, 2006 #1
    the first one says:

    Prove that:
    a) [tex] |sin z | \geq |sin x| [/tex]
    b) [tex] |cos z | \geq |cos x| [/tex]


    Where I guess z = x+iy...

    What I have done:
    Well I am pretty stumped on this one, I have though about expanding sin z into [tex] (sin x) (cosh y) + (i cos x) (sinh y) [/tex]. I dont think that helps me prove anything, but it seems like more terms means it would be greater than just a sin x :uhh:




    Second problemo:

    We see the anuglar momentum components
    [tex] (L_x - i L_y) != (L_x +iL_y)* [/tex]


    Gosh I've tried a lot on this one, I really don't know too too much about QM, so its been tough. It just seems to go against the definition of a conjugate, so I dunno....
     
  2. jcsd
  3. Feb 10, 2006 #2

    Tide

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    First:

    Think a little harder! :)

    HINT: What is the absolute value of the expression?

    Second: What exactly does that exclamation point mean?
     
  4. Feb 10, 2006 #3
    oh i didnt know how to put the does not equal to sign, so I put the !=.
     
  5. Feb 10, 2006 #4
    hmmmm well of course I know |z| = [tex]\sqrt{x^2 +y^2}[/tex]

    so would |sin z| = [tex] \sqrt{(sin x)^2 (cosh y)^2 + (i cos x)^2 (sinh y)^2} [/tex]
     
    Last edited: Feb 10, 2006
  6. Feb 10, 2006 #5

    Tide

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    Not quite. You need to multiply sin z by its complex conjugate, i.e. the i under the radical doesn't belong there.
     
  7. Feb 10, 2006 #6
    hmm ok so I get[tex] (sin^2 x) (cosh^2y) + (cos^2 x) (sinh^2 y) [/tex]

    but that gives me [tex]|sin z|^2[/tex], and then hmm, I need to take the sqaure root of that to get get back to |sin z|?
     
  8. Feb 10, 2006 #7

    Tom Mattson

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    I think you'll find the identity [itex]\cos^2(x)=1-\sin^2(x)[/itex] quite helpful in finishing this off.

    As for the QM question I think what they're driving at here is that you don't take the complex conjugate of operators. Instead you take the Hermitian conjugate. That means that [itex]L_x-iL_y=(L_x+iL_y)^{\dagger}[/itex].

    And by the way the not-equal-to symbol is given by [itex]\neq[/itex] (click the image to see the code).
     
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