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Math Help - Imaginary Part

  1. #1
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    Imaginary Part

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    Show that Im(2 + z +3z^4) <= 4 when |z|<=1
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    The regular way is to go |2 + z + 3z^4| <= 2 + |z| + 3|z|^4
    then we know |z|=1 so 2+1+3 =6 and we want <=4, so im missing some sort of identity or substitution .
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scubasteve123 View Post
    _
    Show that Im(2 + z +3z^4) <= 4 when |z|<=1
    _
    The regular way is to go |2 + z + 3z^4| <= 2 + |z| + 3|z|^4
    then we know |z|=1 so 2+1+3 =6 and we want <=4, so im missing some sort of identity or substitution .
    I think you are misinterpreting the question. It asks for \Im\left(2+z+3z^4\right)\leqslant4 NOT \left|2+z+3z^4\right|\leqslant 4
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I think you are misinterpreting the question. It asks for \Im\left(2+z+3z^4\right)\leqslant4 NOT \left|2+z+3z^4\right|\leqslant 4
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    okay but I have another practice question where |Re(2+z +z^3)|<=4 and thats the method that was used there so I thought it would be similar but i suppose not. Are there any ideas that would help me?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scubasteve123 View Post
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    okay but I have another practice question where |Re(2+z +z^3)|<=4 and thats the method that was used there so I thought it would be similar but i suppose not. Are there any ideas that would help me?
    \left|\Im(z)\right|\leqslant |z| but that doesn't help here, does it?
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    \left|\Im(z)\right|\leqslant |z| but that doesn't help here, does it?
    No I found that property also and cant find much use unfortunately
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scubasteve123 View Post
    No I found that property also and cant find much use unfortunately
    So how do you think we should go about doing this?
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  7. #7
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    It is given that mod z = 1
    that means that
    x^2 + y^2 = 1

    Let z = X + iy
    subtitute this into the given expression, and then take the imaginary part. Substitute x^2 + y^2 =1 into the imaginary part wherever possible
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  8. #8
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    Hopefully this is right....

    Im(2 + z +3z^4)

    let, z= rcosX+isinX
    z^n = r^n {cos(nX)+isin(nX)} by de moivre

    Im(2 + z +3z^4) = Im(2 + r{cosX+isinX} + 3r^4 cos(4X)+isin(4X))
    = rsinX + 3r^4(sin{4X})

    but |z|<=1, so r<=1 for all X

    so Im(2 + z +3z^4) <= 1 + 3 = 4

    QED
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