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Thread: Here is a fun problem.

  1. #1
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    Here is a fun problem.

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

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


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

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

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

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

    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 $\displaystyle z~\&~w$ are complex, we cannot write $\displaystyle z<w$.
    The complex numbers are not ordered.
    Therefore, $\displaystyle e^{i\alpha } \leqslant e^{i\beta } $ makes no sense whatsoever.

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

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

    is this equivalent to:

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

    does that make sense?
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  4. #4
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    Quote Originally Posted by MichaelMath View Post
    $\displaystyle d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta})$
    is this equivalent to:
    $\displaystyle ||z|e^{i \theta} - e^{i \theta}| \, \leq \, ||z|e^{i \theta} - e^{i \alpha}| \, \leq \, ||z|e^{i \theta} + e^{i \theta}|$ ?
    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 $\displaystyle \left| {\left| z \right|e^{i\phi } - e^{i\alpha } } \right| \leqslant \left| z \right| + 1$.
    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 $\displaystyle (X,d)$ be the metric space $\displaystyle \mathbb{C}$ with its usual metric: $\displaystyle d(z,w)=|z-w|$. Prove the following and explain it geometrically:

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

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


    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 $\displaystyle z=|z|e^{i\theta}$ where $\displaystyle \theta=\text{Arg}(z)$.
    The points $\displaystyle e^{i\alpha} $ are on the unit circle for any $\displaystyle \alpha \in \mathbb{R}$.
    The line determined by $\displaystyle z~\&~(0,0)$ intersects the unit circle at $\displaystyle e^{i\theta}~\&~ -e^{i\theta} $.

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