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Math Help - Let w be a complex number with 0 < |w| < 1. Show that the set of all z...

  1. #1
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    Let w be a complex number with 0 < |w| < 1. Show that the set of all z...

    Let w be a complex number with 0 < |w| < 1. Show that the set of all z with these properties is as described

    1: Show that |z - w| < |1 - w*z| is the disc {z:|z| < 1}
    2: Show that |z - w| = |1 - w*z| is the circle {z:|z| = 1}
    3: Show that |z - w| > |1 - w*z| is the circle {z:|z| > 1}

    I squared both sides and simplified... assuming that it would equal

    a^2 + b^2 < 1

    given z= a +bi and w= c + di

    but i got

    a^2 + b^2 + c^2 + d^2 - (a^2)(c^2) - (b^2)(c^2) - (b^2)(d^2) -(a^2)(d^2) < 1

    which i simplified to...

    (|z|^2)(1 - (|w|^2)) + |w|^2 < 1

    anyone know what i am doing wrong?
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  2. #2
    Junior Member nimon's Avatar
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    So close!

    Don't kick yourself, but you've already done all of the hard work! Assuming that your working is correct, the answer lies in a simple reaarangement to make |z| the subject of

    |z|^{2}(1-|w|^{2})+|w|^{2}<1.


    Note that when rearranging you will be dividing by a possibly very small number. Can you state which condition allows you to do this without making a division by 0 error?
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  3. #3
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    Quote Originally Posted by nimon View Post
    Don't kick yourself, but you've already done all of the hard work! Assuming that your working is correct, the answer lies in a simple reaarangement to make |z| the subject of

    |z|^{2}(1-|w|^{2})+|w|^{2}<1.



    Note that when rearranging you will be dividing by a possibly very small number. Can you state which condition allows you to do this without making a division by 0 error?
    oook so i can jsut move |w|^2 over to the right and divide to get 1 on the right side?
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  4. #4
    Junior Member nimon's Avatar
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    Bingo!
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  5. #5
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    Wow I am so thick in the head sometimes
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