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Math Help - Big Problem

  1. #1
    Bertie
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    Big Problem

    Can anyone help me do the following problem. Have been trying for ages but cant prove the answer.

    let z=re^ih (where h is not fixed)

    show that if 0<h<(Pi/2) then mod of cosz tends to infinity as r tends to infinity.

    what is the behaviour of mod cosz as r tends to infinity and h=(pi/2)
    (sorry i cant find the symbols for pi!)

    thanks very much
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Bertie View Post
    Can anyone help me do the following problem. Have been trying for ages but cant prove the answer.

    let z=re^ih (where h is not fixed)

    show that if 0<h<(Pi/2) then mod of cosz tends to infinity as r tends to infinity.

    what is the behaviour of mod cosz as r tends to infinity and h=(pi/2)
    (sorry i cant find the symbols for pi!)

    thanks very much
    What you need to know is that for a general complex number:

    cos(z) = [e(iz)+e(-iz)]/2.

    Write out your z in Cartesian form (like a+ib) and substitute it in this
    and the answer should follow.

    RonL
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  3. #3
    Grand Panjandrum
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    A similar question is:

    Quote Originally Posted by edgar davids
    sorry to keep hassling you all the time but youre the one who seems to be the most helpful and the mostable to answer all my problems!

    can you give me a hand with part a and b of this question please.

    thanks in advance
    Edgar


    Let z = re^itheta, where theta is fixed.

    (a) Show that if 0 < theta< pi/2, then | cos z| tends to infinity as r tends to infinity.
    (b) Describe the behaviour of | cos z| as r tends to infinity when theta = pi/2
    (a) Assume r and theta are real.

    then cos(z)=[e^(iz) + e^(-iz)]/2

    Write z=r (cos(theta)+isin(theta)), then

    e^(iz)=e^(r i cos(theta)) e^(- r sin(theta))

    Therefore for theta in the range specified:

    |e^(iz)| = |e^(- r sin(theta))|

    as |e^(r i cos(theta))|=1.

    similarly:

    |e^(-iz)| = |e^(r sin(theta))|.

    Now

    |cos(z)|=|[e^(iz) + e^(-iz)]/2|

    ...........>= ||e^(iz)/2| - |e^(-iz)]/2||.

    ...........>= |e^(- r sin(theta))|/2 - |e^(r sin(theta))|/2

    but as r goes to infinity the second of the terms on the right goes to infinity, while the second goes to 0.

    so |cos(z)| goes to infinity.

    (b) When theta=pi/2 the cos(theta) terms are all zero and the sin(theta) terms all equal 1, so:

    cos(z) = [e^(- r) + e( r)]/2 , and so |cos(z)| goes to infinity as r goes to infinity.


    RonL
    Last edited by CaptainBlack; October 25th 2006 at 08:26 AM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    A similar question is:



    (a) Assume r and theta are real.

    then cos(z)=[e^(iz) + e^(-iz)]/2

    Write z=r (cos(theta)+isin(theta)), then

    e^(iz)=e^(r cos(theta)) e^(r i sin(theta))

    Therefore for theta in the range specified:

    |e^(iz)| = |e^(r cos(theta))|
    There is a slight problem here, but it doesn't affect the result.

    e^{iz} = e^{ir \, cos \theta} e^{i^2r \, sin \theta} = e^{-r sin \theta}e^{ir \, cos \theta}

    So the real part of e^(iz) is e^(-r sin(theta)), not the cosine version. Additionally this means that |e^(iz)| = |e^(-r sin(theta))| -> 0 as r -> infinity, so it is the |e^(-iz)| term that dominates as r -> infinity. But as this term also goes to infinity, we are merely switching the dominant term and the proof goes through otherwise unaffected.

    -Dan
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    There is a slight problem here, but it doesn't affect the result.

    e^{iz} = e^{ir \, cos \theta} e^{i^2r \, sin \theta} = e^{-r sin \theta}e^{ir \, cos \theta}

    So the real part of e^(iz) is e^(-r sin(theta)), not the cosine version. Additionally this means that |e^(iz)| = |e^(-r sin(theta))| -> 0 as r -> infinity, so it is the |e^(-iz)| term that dominates as r -> infinity. But as this term also goes to infinity, we are merely switching the dominant term and the proof goes through otherwise unaffected.

    -Dan
    Opps

    But it does change the second part, which I have adjusted accordingly.
    The attachment shows the result of a numerical experiment to check the
    result.

    RonL
    Attached Thumbnails Attached Thumbnails Big Problem-gash.jpg  
    Last edited by CaptainBlack; October 25th 2006 at 08:29 AM.
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