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Math Help - Here is a fun problem.

  1. #1
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    Here is a fun problem.

    (X,d) is the metric space \mathbb{C} with metric d(z,w) = |z-w|. Prove:

    d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta}) , for all \alpha \in \mathbb{R} , where z=|z|e^{i \theta}


    Not sure what's going on here so I started going backwards for the hell of it:

    |z|e^{i \theta} - e^{i \theta} \leq |z|e^{i \theta} - e^{i \alpha} \leq |z|e^{i \theta} + e^{i \theta}

    -e^{i \theta} \leq -e^{i \alpha} \leq  e^{i \theta}

    |-e^{i \alpha}| \leq  e^{i \theta}

    Is this correct or useful in any way? Can somebody explain this geometrically?



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  2. #2
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    You need to review your post because it contains many mistakes.
    If z~\&~w are complex, we cannot write z<w.
    The complex numbers are not ordered.
    Therefore, e^{i\alpha }  \leqslant e^{i\beta } makes no sense whatsoever.

    Moreover, do you understand that \left( {\forall t \in \mathbb{R}} \right)\left[ {\left| {e^{it} } \right| = \left| { - e^{it} } \right| = 1} \right]?

    This is true \left| {\left| z \right|e^{i\phi }  - e^{i\alpha } } \right| \leqslant \left| z \right| + 1.
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  3. #3
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    d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta})

    is this equivalent to:

    ||z|e^{i \theta} - e^{i \theta}| \, \leq \, ||z|e^{i \theta} - e^{i \alpha}|  \, \leq \, ||z|e^{i \theta} + e^{i \theta}| ?

    does that make sense?
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  4. #4
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    Quote Originally Posted by MichaelMath View Post
    d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta})
    is this equivalent to:
    ||z|e^{i \theta} - e^{i \theta}| \, \leq \, ||z|e^{i \theta} - e^{i \alpha}|  \, \leq \, ||z|e^{i \theta} + e^{i \theta}| ?
    Yes those two are equivalent.
    But that proves nothing.
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  5. #5
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    Quote Originally Posted by Plato View Post
    This is true \left| {\left| z \right|e^{i\phi }  - e^{i\alpha } } \right| \leqslant \left| z \right| + 1.
    I guess you were hinting that I should start here?
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  6. #6
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    Quote Originally Posted by MichaelMath View Post
    I guess you were hinting that I should start here?
    Actually, I am not hinting at anything.
    The question makes no sense to me.
    Did you copy it correctly?
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  7. #7
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    Let (X,d) be the metric space \mathbb{C} with its usual metric: d(z,w)=|z-w|. Prove the following and explain it geometrically:

    d(z, e^{i \theta}) \leq d(z, e^{i \alpha}) \leq d(z,-e^{i \theta})

    for all \alpha \in \mathbb(R) where z=|z|e^{i \theta}.


    Any improvement?
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  8. #8
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    Well that does clear away the clutter.
    Any complex number can be written as z=|z|e^{i\theta} where \theta=\text{Arg}(z).
    The points  e^{i\alpha} are on the unit circle for any \alpha \in \mathbb{R}.
    The line determined by z~\&~(0,0) intersects the unit circle at  e^{i\theta}~\&~ -e^{i\theta} .

    So the real numbers d\left(z,e^{i\theta}\right)~\&~ d\left(z,-e^{i\theta}) represent the distances of points on the unit circle at minimum and maximum from z.
    Last edited by Plato; October 22nd 2010 at 04:17 AM.
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  9. #9
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    Is there an i missing in the last part?
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  10. #10
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    Quote Originally Posted by Plato View Post
    So the real numbers d\left(z,e^{i\theta}\right)~\&~ d\left(z,-e^{i\theta}) represent the distances of points on the unit circle at minimum and maximum from z.
    Does this last part imply that d(z, any other point on the unit circle ) must be equal to or lie between d(z,e^{i \theta}) and d(z,-e^{i \theta}) ?
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  11. #11
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    Quote Originally Posted by MichaelMath View Post
    Does this last part imply that d(z, any other point on the unit circle ) must be equal to or lie between d(z,e^{i \theta}) and d(z,-e^{i \theta}) ?
    It means exactly that. If \text{Arg}(z)=\theta then d\left(z,e^{\theta i }\right)\le d\left(z,e^{t i }\right)\le d\left(z,-e^{\theta i }\right) where t\in \mathbb{R}.
    Last edited by Plato; October 26th 2010 at 09:36 AM.
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