Thread: Here is a fun problem.

1. 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?

2. You need to review your post because it contains many mistakes.
If $z~\&~w$ are complex, we cannot write $z.
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 $\left| {\left| z \right|e^{i\phi } - e^{i\alpha } } \right| \leqslant \left| z \right| + 1$.

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

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?

4. Originally Posted by MichaelMath
$d(z,e^{i \theta}) \leq d(z,e^{i \alpha}) \leq d(z,-e^{i \theta})$
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.

5. Originally Posted by Plato
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?

6. Originally Posted by MichaelMath
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?

7. 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?

8. 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 $z$.

9. Is there an $i$ missing in the last part?

10. Originally Posted by Plato
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 $d(z,-e^{i \theta})$ ?

11. Originally Posted by MichaelMath
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 $t\in \mathbb{R}$.