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Show that sin z

  1. Mar 9, 2009 #1
    hi to all
    can i obtain the answer of this problem :
    show that :

    modulus of (sin z ) > = modulus of (sin X )

    where z = x + i Y
     
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 9, 2009 #2

    Integral

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    What do you know about the modulus of complex numbers?

    Note that on these forums we do not do proplems for you, we help, YOU do them.
     
  4. Mar 9, 2009 #3
    also i am asking about this problem :

    what part of z - plane corresponds to the interior of the unit circle in the w-plane if
    a) -w = (z-1)/(z+1)
    b) -w = (z - i)/(z + i)
     
  5. Mar 9, 2009 #4
    i can say that
    (modulus fo sin z )^2 = sin z * (sin z )*
     
  6. Mar 9, 2009 #5

    Integral

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    One proplem per thread please. Now what has your first post got to do with sin(z)?
     
  7. Mar 9, 2009 #6
    really i am sorry ..

    sin z = ( exp i(x+iy) - exp -i(x+iy))/2i
    ( sin z )* = exp ( -i(x-iy) - exp ( i(x-iy)/(-2i)

    (mod. (sin z) )^2 = ( exp i(x+iy) - exp -i(x+iy))/2i * exp ( -i(x-iy) - exp ( i(x-iy)/(-2i)

    is this step good ?
     
  8. Mar 9, 2009 #7

    Office_Shredder

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    Your original post is asking about the modulus of a complex number; no question here involves sin(z). Unless you specify what you're trying to answer we can't help
     
  9. Mar 9, 2009 #8
    agin i am sorry ... i mean sin z
    and sin x

    thank u
     
  10. Mar 11, 2009 #9
    no one will try with me !!!!!!!

    ok
    i say : sin z = sin ( x + iy ) = sin x cos iy + cos x sin iy
    = sin x cosh y - i cos x sinh y

    modulus (sin z) > = modulus ( sin x cosh y) - modulus ( i cos x sinh y)

    is this trueeeeeeeeee
     
  11. Mar 11, 2009 #10
    where r uuuuuuuuuuu
     
  12. Mar 11, 2009 #11

    Dick

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    That much is true. The rest isn't. Find the square of the modulus of sin(z) by multiplying sin(z) by it's complex conjugate.
     
  13. Mar 13, 2009 #12
    thanx ... a lot
    i tried and wrote the following ..

    Sin ( z ) = sin ( x + iy )
    = sin x cos iy + sin iy cos x
    = sin x cosh y + i sinh y cos x

    Now
    │sin z │2 = ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 (cos x )2
    = ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 (1 – ( sin x ) 2)
    = ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 - ( sinh y)2 ( sin x ) 2
    = ( sin x ) 2 (( cosh y ) 2 - ( sinh y)2 ) + ( sinh y)2
    so , we have

    │sin z │2 = ( sin x ) 2 + ( sinh y)2

    but the problem of y variable still hold !!!!!

    can i say that :
    ( sin x ) 2 + ( sinh y)2 >= ( sin x ) 2 ??????
     
  14. Mar 13, 2009 #13

    Dick

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    Sure you can. sinh(y)^2>=0. Just because it's the square of something.
     
  15. Mar 13, 2009 #14
    so , then i can say that

    │sin z │2 >= ( sin x ) 2

    taking square root to both sides gives

    │sin z │>= │sin x│

    did it finish ???
     
  16. Mar 13, 2009 #15

    Dick

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    That's fine. But you don't sound very sure of yourself?
     
  17. Mar 13, 2009 #16
    may be .... thank you very much ...
     
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