Proving that the triangle inequality holds for a metric on $mathbbC$

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Show that $(X,d)$ is a metric space where $X =Bbb C $ and the distance function is defined as:
$$d(x,y) = frac x-ysqrt ^2 + sqrt 1 + , text for x,y in Bbb C.$$

I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.

inequality metric-spaces

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edited Jan 17 '15 at 13:13

user169373

asked Jan 14 '15 at 16:01

user8795

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  If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
  – Scientifica
  Jan 15 '15 at 13:06
  

  
  
  
  


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  Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
  – Ivo Terek
  Jan 16 '15 at 14:02
  

  
  
  
  


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  @Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
  – Alex Ravsky
  Jan 17 '15 at 15:04
  

  
  
  
  


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  I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
  – Ivo Terek
  Jan 17 '15 at 16:33
  
  

  
  
  
  


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  @AlexRavsky thanks for your comment.
  – Scientifica
  Jan 18 '15 at 10:08
  

  
  
  
  

 | 
show 2 more comments

up vote
23
down vote

favorite

20

Show that $(X,d)$ is a metric space where $X =Bbb C $ and the distance function is defined as:
$$d(x,y) = frac x-ysqrt ^2 + sqrt 1 + , text for x,y in Bbb C.$$

I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.

inequality metric-spaces

share|cite|improve this question

edited Jan 17 '15 at 13:13

user169373

asked Jan 14 '15 at 16:01

user8795

5,25261842

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
  – Scientifica
  Jan 15 '15 at 13:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
  – Ivo Terek
  Jan 16 '15 at 14:02
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
  – Alex Ravsky
  Jan 17 '15 at 15:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
  – Ivo Terek
  Jan 17 '15 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexRavsky thanks for your comment.
  – Scientifica
  Jan 18 '15 at 10:08
  

  
  
  
  

 | 
show 2 more comments

up vote
23
down vote

favorite

20

up vote
23
down vote

favorite

20

20

Show that $(X,d)$ is a metric space where $X =Bbb C $ and the distance function is defined as:
$$d(x,y) = frac x-ysqrt ^2 + sqrt 1 + , text for x,y in Bbb C.$$

I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.

inequality metric-spaces

share|cite|improve this question

edited Jan 17 '15 at 13:13

user169373

asked Jan 14 '15 at 16:01

user8795

5,25261842

Show that $(X,d)$ is a metric space where $X =Bbb C $ and the distance function is defined as:
$$d(x,y) = frac x-ysqrt ^2 + sqrt 1 + , text for x,y in Bbb C.$$

I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.

inequality metric-spaces

share|cite|improve this question

edited Jan 17 '15 at 13:13

user169373

asked Jan 14 '15 at 16:01

user8795

5,25261842

share|cite|improve this question

share|cite|improve this question

edited Jan 17 '15 at 13:13

user169373

edited Jan 17 '15 at 13:13

user169373

edited Jan 17 '15 at 13:13

user169373

asked Jan 14 '15 at 16:01

user8795

5,25261842

asked Jan 14 '15 at 16:01

user8795

5,25261842

asked Jan 14 '15 at 16:01

user8795

5,25261842

5,25261842

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
  – Scientifica
  Jan 15 '15 at 13:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
  – Ivo Terek
  Jan 16 '15 at 14:02
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
  – Alex Ravsky
  Jan 17 '15 at 15:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
  – Ivo Terek
  Jan 17 '15 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexRavsky thanks for your comment.
  – Scientifica
  Jan 18 '15 at 10:08
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
  – Scientifica
  Jan 15 '15 at 13:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
  – Ivo Terek
  Jan 16 '15 at 14:02
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
  – Alex Ravsky
  Jan 17 '15 at 15:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
  – Ivo Terek
  Jan 17 '15 at 16:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexRavsky thanks for your comment.
  – Scientifica
  Jan 18 '15 at 10:08
  

  
  
  
  

2

2

If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
– Scientifica
Jan 15 '15 at 13:06

If this inequality holds: $forall x,y,zinmathbbC,,|x^2(y-z)|+|y^2(x-z)|ge |z^2(x-y)|$ then I can answer your question.
– Scientifica
Jan 15 '15 at 13:06

2

2

Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
– Ivo Terek
Jan 16 '15 at 14:02

Recently showed up a duplicate of this. None can be closed since there are no answers. Keep an eye there, something might show up. I tried a few things here myself, but I couldn't get anything either.
– Ivo Terek
Jan 16 '15 at 14:02

2

2

@Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
– Alex Ravsky
Jan 17 '15 at 15:04

@Scientifica If in your inequality we put $x=0$ then it will transform to a false one: $|y^2z|ge |z^2y|.$
– Alex Ravsky
Jan 17 '15 at 15:04

1

1

I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
– Ivo Terek
Jan 17 '15 at 16:33

I posted this problem here, and someone had the ideia of finding a metric in $S^2$ which induces this one... (it's in portuguese, but the idea is mainly what I said... and no one will have trouble with formulas, I suppose)
– Ivo Terek
Jan 17 '15 at 16:33

@AlexRavsky thanks for your comment.
– Scientifica
Jan 18 '15 at 10:08

@AlexRavsky thanks for your comment.
– Scientifica
Jan 18 '15 at 10:08

 | 
show 2 more comments

3 Answers
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1
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A partial result. Let $x,y,zinmathbbC$ such that $|y|leqmin,$ then
$$d(x,z)=frac2sqrt^2+sqrt^2leqfracsqrt^2+sqrt^2$$$$=fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|leqmin,$ then
$$sqrt^2+sqrtyleqsqrt^2+sqrt^2$$
and
$$sqrty+sqrt^2leqsqrt^2+sqrt^2$$
which together with the result above on the main metric $d(x,z)$ yield
$$fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$$$leqfracx-ysqrt^2+sqrty+fracy-zsqrty+sqrt^2=d(x,y)+d(y,z)$$
In other words
$$d(x,z)leq d(x,y)+d(y,z)$$

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answered Jan 21 '15 at 15:29

Arian

5,235817

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  Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
  – user2345215
  Jan 22 '15 at 14:08
  

  
  
  
  


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  @user2345215: well this is just a partial result. It is true that the inequality is more subtle.
  – Arian
  Jan 23 '15 at 13:02
  

  
  
  
  

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This migth be helpful: I'll proof that the function:
$$d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $$ is a metric in the complex numbers. My idea to prove it is to use the stereographic projection of $mathbbC$ into $S^2$ (the unit sphere of $mathbbR^3$) and define the metric $d$ via the euclidean metric in $mathbbR^3$. The technical details are the following:

Let $z=a+ib$ a complex number, then, the stereographic projection is a function $H:mathbbCtoS^2subsetmathbbR^3$ defined by:
$$H(z)=Hleft( a + ib right) = left( ta,tb,1 - t right)quad,quad;t = frac2left^2 + 1$$
The stereographic projection is an inyective function

For any two complex numbers $z_0=a_0+ib_0$ and $z_1=a_1+ib_1$ define a function $rho$ in the following way:
$$rho left( z_0,z_1 right) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right)$$
Is easy to verify that $rho$ is a metric in $mathbbC$:

• $rho (z_0,z_1) = 0 Leftrightarrow d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = 0 Leftrightarrow Hleft( z_0 right) = Hleft( z_1 right) Leftrightarrow z_0 = z_1$


• $rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = d_mathbbR^3left( Hleft( z_1 right),Hleft( z_0 right) right) = rho (z_1,z_0)$


• 
  

  Let $z_2$ be another complex number, then,
  $$rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) le d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right)$$
  (the inequality is given by the triangular inequality of $d_mathbbR^3$).

  
  
  

  By the definition of $rho$ we have $d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right) = rho (z_0,z_1) + rho (z_1,z_2)$, so $$rho (z_0,z_1) le rho (z_0,z_1) + rho (z_1,z_2).$$

  
  

Now, the ``only'' thing left to prove is that $rho (z_0,z_1) = d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $. This demonstration is not very interesnting (it's just a lot of algebraic manipulations). Here goes the details:
beginalign*
rho (z_0,z_1) &= sqrt left( t_0a_0 - t_1a_1 right)^2 + left( t_0b_0 - t_1b_1 right)^2 + left( t_1 - t_0 right)^2nonumber\
&= sqrt left( t_0a_0 right)^2 - 2t_0t_1a_0a_1 + left( t_1a_1 right)^2 + left( t_0b_0 right)^2 - 2t_0t_1b_0b_1 + left( t_1b_1 right)^2 + t_1^2 - 2t_0t_1 + t_0^2nonumber\
&= sqrt t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)
endalign*
Let $A = t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right)$, $B = t_0^2left( a_0^2 + b_0^2 + 1 right)$ and $C = t_1^2left( a_1^2 + b_1^2 + 1 right)$, so $$rho left( z_0,z_1 right) = sqrt A + B + C$$

Now, note that
beginalign*
frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 &= left| z_0 - z_1 right|sqrt frac21 + ^2 sqrt frac21 + z_1 right^2 = sqrt left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 sqrt t_0t_1nonumber\
&= sqrt t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)
endalign*
Set $D=t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)$, so $$d(z_0,z_1)=sqrtD$$
Now,
beginalign*
A &= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) nonumber\
&= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 + a_o^2 - a_0^2 + a_1^2 - a_1^2 + b_o^2 - b_0^2 + b_1^2 - b_1^2 right)nonumber\
&= t_0t_1left( a_o^2 - 2a_0a_1 + a_1^2 + b_o^2 - 2b_ob_1 + b_1^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 right) - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)nonumber\
&= D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)
endalign*
Putting this in $rho left( z_0,z_1 right) = sqrt A + B + C$ we have
$$rho left( z_0,z_1 right) = sqrt D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C$$
so, we just need to proof that $E= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C $ equals zero.

Now,
beginalign*
E &= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)nonumber\
&= a_0^2left( - t_0t_1 + t_0^2 right) + a_1^2left( - t_0t_1 + t_1^2 right) + b_0^2left( - t_0t_1 + t_0^2 right) + b_1^2left( - t_0t_1 + t_1^2 right) + left( t_0 - t_1 right)^2nonumber\
&= a_0^2t_0left( - t_1 + t_0 right) + a_1^2t_1left( - t_0 + t_1 right) + b_0^2t_0left( - t_1 + t_0 right) + b_1^2t_1left( - t_0 + t_1 right) + left( t_0 - t_1 right)^2\
&= left( t_0 - t_1 right)left( a_0^2t_0 - a_1^2t_1 + b_0^2t_0 - b_1^2t_1 + t_0 - t_1 right)\
&= left( t_0 - t_1 right)Bigl( t_0left( a_0^2 + b_0^2 right) - t_1left( a_1^2 + b_1^2 right) + t_0 - t_1 Bigr) \
&= left( t_0 - t_1 right)left( t_0^2 - t_1 z_1 right^2 + t_0 - t_1 right) \
&= left( t_0 - t_1 right)left( underbrace t_0left( ^2 + 1 right)_2 - underbrace t_1left( z_1 right^2 + 1 right)_2 right)\
&= 0
endalign*
So, $rho(z_0,z_1)=d(z_0,z_1)$ and the proof is complete!

Some cool facts about this question:

• The distance $d$ is known as the textbfchordal metric and is interpreted as the euclidean distance between the stereographic projections of two complex numbers. That's why it's called chordal metric (the distance between two points in an sphere is the length of the chord that has the points as its extremes).


• The metric $d$ can be extended to metric in $mathbbC^*$ by setting
  $$dleft( z,infty right) = frac2sqrt 1 + left^2 $$


• $d$ (and its extention) is bounded: the distance between two points is always equal or less than 2.

You can find more about this metric (but not the proof that it's actually a matric) in the books Introduction to Complex Analysis by B.V. Shabat (pages 6 and 7) and Complex Analysis by T.W. Gamelin (section I.3 and especially exercise 6)

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edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

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  Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
  – Martin R
  Jun 26 '15 at 17:55
  
  

  
  
  
  


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  Oh, you're totally right
  – Charly
  Jul 1 '15 at 15:58
  

  
  
  
  

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I did a small step before I stuck, but it may be useful for others. So, it seems the following.

The cases $|y|lemin,$ and $|y|gemax,$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^-1$, $v=y^-1$, and $w=z^-1$).

So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics.

Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=sqrt ^2 + sqrt 1 + $, and $v_2=sqrt ^2 + sqrt 1 + $ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds
$$fraccosangle oyx-cosangle zyo=frac v_1v_2.$$

But $$cosangle oyx=frac(y-x,y),$$
$$- cosangle zyo=frac(y-z,y)y.$$

So $$frac(y-x,y)=
frac(y-z,y)left(sqrt y + sqrt ^2right).$$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

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3 Answers
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3 Answers
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up vote
1
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A partial result. Let $x,y,zinmathbbC$ such that $|y|leqmin,$ then
$$d(x,z)=frac2sqrt^2+sqrt^2leqfracsqrt^2+sqrt^2$$$$=fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|leqmin,$ then
$$sqrt^2+sqrtyleqsqrt^2+sqrt^2$$
and
$$sqrty+sqrt^2leqsqrt^2+sqrt^2$$
which together with the result above on the main metric $d(x,z)$ yield
$$fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$$$leqfracx-ysqrt^2+sqrty+fracy-zsqrty+sqrt^2=d(x,y)+d(y,z)$$
In other words
$$d(x,z)leq d(x,y)+d(y,z)$$

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answered Jan 21 '15 at 15:29

Arian

5,235817

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
  – user2345215
  Jan 22 '15 at 14:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2345215: well this is just a partial result. It is true that the inequality is more subtle.
  – Arian
  Jan 23 '15 at 13:02
  

  
  
  
  

add a comment | 

up vote
1
down vote

A partial result. Let $x,y,zinmathbbC$ such that $|y|leqmin,$ then
$$d(x,z)=frac2sqrt^2+sqrt^2leqfracsqrt^2+sqrt^2$$$$=fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|leqmin,$ then
$$sqrt^2+sqrtyleqsqrt^2+sqrt^2$$
and
$$sqrty+sqrt^2leqsqrt^2+sqrt^2$$
which together with the result above on the main metric $d(x,z)$ yield
$$fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$$$leqfracx-ysqrt^2+sqrty+fracy-zsqrty+sqrt^2=d(x,y)+d(y,z)$$
In other words
$$d(x,z)leq d(x,y)+d(y,z)$$

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answered Jan 21 '15 at 15:29

Arian

5,235817

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
  – user2345215
  Jan 22 '15 at 14:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2345215: well this is just a partial result. It is true that the inequality is more subtle.
  – Arian
  Jan 23 '15 at 13:02
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

A partial result. Let $x,y,zinmathbbC$ such that $|y|leqmin,$ then
$$d(x,z)=frac2sqrt^2+sqrt^2leqfracsqrt^2+sqrt^2$$$$=fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|leqmin,$ then
$$sqrt^2+sqrtyleqsqrt^2+sqrt^2$$
and
$$sqrty+sqrt^2leqsqrt^2+sqrt^2$$
which together with the result above on the main metric $d(x,z)$ yield
$$fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$$$leqfracx-ysqrt^2+sqrty+fracy-zsqrty+sqrt^2=d(x,y)+d(y,z)$$
In other words
$$d(x,z)leq d(x,y)+d(y,z)$$

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answered Jan 21 '15 at 15:29

Arian

5,235817

A partial result. Let $x,y,zinmathbbC$ such that $|y|leqmin,$ then
$$d(x,z)=frac2sqrt^2+sqrt^2leqfracsqrt^2+sqrt^2$$$$=fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$
(Note that the last result is obtained by the triangle inequality on the metric $d^*(x,y)=|x-y|$)
Since $|y|leqmin,$ then
$$sqrt^2+sqrtyleqsqrt^2+sqrt^2$$
and
$$sqrty+sqrt^2leqsqrt^2+sqrt^2$$
which together with the result above on the main metric $d(x,z)$ yield
$$fracx-ysqrt^2+sqrt^2+fracy-zsqrt^2+sqrt^2$$$$leqfracx-ysqrt^2+sqrty+fracy-zsqrty+sqrt^2=d(x,y)+d(y,z)$$
In other words
$$d(x,z)leq d(x,y)+d(y,z)$$

share|cite|improve this answer

answered Jan 21 '15 at 15:29

Arian

5,235817

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 '15 at 15:29

Arian

5,235817

answered Jan 21 '15 at 15:29

Arian

5,235817

answered Jan 21 '15 at 15:29

Arian

5,235817

5,235817

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
  – user2345215
  Jan 22 '15 at 14:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2345215: well this is just a partial result. It is true that the inequality is more subtle.
  – Arian
  Jan 23 '15 at 13:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
  – user2345215
  Jan 22 '15 at 14:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user2345215: well this is just a partial result. It is true that the inequality is more subtle.
  – Arian
  Jan 23 '15 at 13:02
  

  
  
  
  

1

1

Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
– user2345215
Jan 22 '15 at 14:08

Well, this is quite trivial. The evil thing in that in general the inequality doesn't hold if you replace $|x-z|$ by $|x-y|+|y-z|$ right away.
– user2345215
Jan 22 '15 at 14:08

@user2345215: well this is just a partial result. It is true that the inequality is more subtle.
– Arian
Jan 23 '15 at 13:02

@user2345215: well this is just a partial result. It is true that the inequality is more subtle.
– Arian
Jan 23 '15 at 13:02

add a comment | 

up vote
1
down vote

This migth be helpful: I'll proof that the function:
$$d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $$ is a metric in the complex numbers. My idea to prove it is to use the stereographic projection of $mathbbC$ into $S^2$ (the unit sphere of $mathbbR^3$) and define the metric $d$ via the euclidean metric in $mathbbR^3$. The technical details are the following:

Let $z=a+ib$ a complex number, then, the stereographic projection is a function $H:mathbbCtoS^2subsetmathbbR^3$ defined by:
$$H(z)=Hleft( a + ib right) = left( ta,tb,1 - t right)quad,quad;t = frac2left^2 + 1$$
The stereographic projection is an inyective function

For any two complex numbers $z_0=a_0+ib_0$ and $z_1=a_1+ib_1$ define a function $rho$ in the following way:
$$rho left( z_0,z_1 right) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right)$$
Is easy to verify that $rho$ is a metric in $mathbbC$:

• $rho (z_0,z_1) = 0 Leftrightarrow d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = 0 Leftrightarrow Hleft( z_0 right) = Hleft( z_1 right) Leftrightarrow z_0 = z_1$


• $rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = d_mathbbR^3left( Hleft( z_1 right),Hleft( z_0 right) right) = rho (z_1,z_0)$


• 
  

  Let $z_2$ be another complex number, then,
  $$rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) le d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right)$$
  (the inequality is given by the triangular inequality of $d_mathbbR^3$).

  
  
  

  By the definition of $rho$ we have $d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right) = rho (z_0,z_1) + rho (z_1,z_2)$, so $$rho (z_0,z_1) le rho (z_0,z_1) + rho (z_1,z_2).$$

  
  

Now, the ``only'' thing left to prove is that $rho (z_0,z_1) = d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $. This demonstration is not very interesnting (it's just a lot of algebraic manipulations). Here goes the details:
beginalign*
rho (z_0,z_1) &= sqrt left( t_0a_0 - t_1a_1 right)^2 + left( t_0b_0 - t_1b_1 right)^2 + left( t_1 - t_0 right)^2nonumber\
&= sqrt left( t_0a_0 right)^2 - 2t_0t_1a_0a_1 + left( t_1a_1 right)^2 + left( t_0b_0 right)^2 - 2t_0t_1b_0b_1 + left( t_1b_1 right)^2 + t_1^2 - 2t_0t_1 + t_0^2nonumber\
&= sqrt t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)
endalign*
Let $A = t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right)$, $B = t_0^2left( a_0^2 + b_0^2 + 1 right)$ and $C = t_1^2left( a_1^2 + b_1^2 + 1 right)$, so $$rho left( z_0,z_1 right) = sqrt A + B + C$$

Now, note that
beginalign*
frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 &= left| z_0 - z_1 right|sqrt frac21 + ^2 sqrt frac21 + z_1 right^2 = sqrt left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 sqrt t_0t_1nonumber\
&= sqrt t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)
endalign*
Set $D=t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)$, so $$d(z_0,z_1)=sqrtD$$
Now,
beginalign*
A &= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) nonumber\
&= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 + a_o^2 - a_0^2 + a_1^2 - a_1^2 + b_o^2 - b_0^2 + b_1^2 - b_1^2 right)nonumber\
&= t_0t_1left( a_o^2 - 2a_0a_1 + a_1^2 + b_o^2 - 2b_ob_1 + b_1^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 right) - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)nonumber\
&= D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)
endalign*
Putting this in $rho left( z_0,z_1 right) = sqrt A + B + C$ we have
$$rho left( z_0,z_1 right) = sqrt D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C$$
so, we just need to proof that $E= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C $ equals zero.

Now,
beginalign*
E &= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)nonumber\
&= a_0^2left( - t_0t_1 + t_0^2 right) + a_1^2left( - t_0t_1 + t_1^2 right) + b_0^2left( - t_0t_1 + t_0^2 right) + b_1^2left( - t_0t_1 + t_1^2 right) + left( t_0 - t_1 right)^2nonumber\
&= a_0^2t_0left( - t_1 + t_0 right) + a_1^2t_1left( - t_0 + t_1 right) + b_0^2t_0left( - t_1 + t_0 right) + b_1^2t_1left( - t_0 + t_1 right) + left( t_0 - t_1 right)^2\
&= left( t_0 - t_1 right)left( a_0^2t_0 - a_1^2t_1 + b_0^2t_0 - b_1^2t_1 + t_0 - t_1 right)\
&= left( t_0 - t_1 right)Bigl( t_0left( a_0^2 + b_0^2 right) - t_1left( a_1^2 + b_1^2 right) + t_0 - t_1 Bigr) \
&= left( t_0 - t_1 right)left( t_0^2 - t_1 z_1 right^2 + t_0 - t_1 right) \
&= left( t_0 - t_1 right)left( underbrace t_0left( ^2 + 1 right)_2 - underbrace t_1left( z_1 right^2 + 1 right)_2 right)\
&= 0
endalign*
So, $rho(z_0,z_1)=d(z_0,z_1)$ and the proof is complete!

Some cool facts about this question:

• The distance $d$ is known as the textbfchordal metric and is interpreted as the euclidean distance between the stereographic projections of two complex numbers. That's why it's called chordal metric (the distance between two points in an sphere is the length of the chord that has the points as its extremes).


• The metric $d$ can be extended to metric in $mathbbC^*$ by setting
  $$dleft( z,infty right) = frac2sqrt 1 + left^2 $$


• $d$ (and its extention) is bounded: the distance between two points is always equal or less than 2.

You can find more about this metric (but not the proof that it's actually a matric) in the books Introduction to Complex Analysis by B.V. Shabat (pages 6 and 7) and Complex Analysis by T.W. Gamelin (section I.3 and especially exercise 6)

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edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
  – Martin R
  Jun 26 '15 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, you're totally right
  – Charly
  Jul 1 '15 at 15:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

This migth be helpful: I'll proof that the function:
$$d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $$ is a metric in the complex numbers. My idea to prove it is to use the stereographic projection of $mathbbC$ into $S^2$ (the unit sphere of $mathbbR^3$) and define the metric $d$ via the euclidean metric in $mathbbR^3$. The technical details are the following:

Let $z=a+ib$ a complex number, then, the stereographic projection is a function $H:mathbbCtoS^2subsetmathbbR^3$ defined by:
$$H(z)=Hleft( a + ib right) = left( ta,tb,1 - t right)quad,quad;t = frac2left^2 + 1$$
The stereographic projection is an inyective function

For any two complex numbers $z_0=a_0+ib_0$ and $z_1=a_1+ib_1$ define a function $rho$ in the following way:
$$rho left( z_0,z_1 right) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right)$$
Is easy to verify that $rho$ is a metric in $mathbbC$:

• $rho (z_0,z_1) = 0 Leftrightarrow d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = 0 Leftrightarrow Hleft( z_0 right) = Hleft( z_1 right) Leftrightarrow z_0 = z_1$


• $rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = d_mathbbR^3left( Hleft( z_1 right),Hleft( z_0 right) right) = rho (z_1,z_0)$


• 
  

  Let $z_2$ be another complex number, then,
  $$rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) le d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right)$$
  (the inequality is given by the triangular inequality of $d_mathbbR^3$).

  
  
  

  By the definition of $rho$ we have $d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right) = rho (z_0,z_1) + rho (z_1,z_2)$, so $$rho (z_0,z_1) le rho (z_0,z_1) + rho (z_1,z_2).$$

  
  

Now, the ``only'' thing left to prove is that $rho (z_0,z_1) = d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $. This demonstration is not very interesnting (it's just a lot of algebraic manipulations). Here goes the details:
beginalign*
rho (z_0,z_1) &= sqrt left( t_0a_0 - t_1a_1 right)^2 + left( t_0b_0 - t_1b_1 right)^2 + left( t_1 - t_0 right)^2nonumber\
&= sqrt left( t_0a_0 right)^2 - 2t_0t_1a_0a_1 + left( t_1a_1 right)^2 + left( t_0b_0 right)^2 - 2t_0t_1b_0b_1 + left( t_1b_1 right)^2 + t_1^2 - 2t_0t_1 + t_0^2nonumber\
&= sqrt t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)
endalign*
Let $A = t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right)$, $B = t_0^2left( a_0^2 + b_0^2 + 1 right)$ and $C = t_1^2left( a_1^2 + b_1^2 + 1 right)$, so $$rho left( z_0,z_1 right) = sqrt A + B + C$$

Now, note that
beginalign*
frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 &= left| z_0 - z_1 right|sqrt frac21 + ^2 sqrt frac21 + z_1 right^2 = sqrt left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 sqrt t_0t_1nonumber\
&= sqrt t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)
endalign*
Set $D=t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)$, so $$d(z_0,z_1)=sqrtD$$
Now,
beginalign*
A &= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) nonumber\
&= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 + a_o^2 - a_0^2 + a_1^2 - a_1^2 + b_o^2 - b_0^2 + b_1^2 - b_1^2 right)nonumber\
&= t_0t_1left( a_o^2 - 2a_0a_1 + a_1^2 + b_o^2 - 2b_ob_1 + b_1^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 right) - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)nonumber\
&= D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)
endalign*
Putting this in $rho left( z_0,z_1 right) = sqrt A + B + C$ we have
$$rho left( z_0,z_1 right) = sqrt D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C$$
so, we just need to proof that $E= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C $ equals zero.

Now,
beginalign*
E &= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)nonumber\
&= a_0^2left( - t_0t_1 + t_0^2 right) + a_1^2left( - t_0t_1 + t_1^2 right) + b_0^2left( - t_0t_1 + t_0^2 right) + b_1^2left( - t_0t_1 + t_1^2 right) + left( t_0 - t_1 right)^2nonumber\
&= a_0^2t_0left( - t_1 + t_0 right) + a_1^2t_1left( - t_0 + t_1 right) + b_0^2t_0left( - t_1 + t_0 right) + b_1^2t_1left( - t_0 + t_1 right) + left( t_0 - t_1 right)^2\
&= left( t_0 - t_1 right)left( a_0^2t_0 - a_1^2t_1 + b_0^2t_0 - b_1^2t_1 + t_0 - t_1 right)\
&= left( t_0 - t_1 right)Bigl( t_0left( a_0^2 + b_0^2 right) - t_1left( a_1^2 + b_1^2 right) + t_0 - t_1 Bigr) \
&= left( t_0 - t_1 right)left( t_0^2 - t_1 z_1 right^2 + t_0 - t_1 right) \
&= left( t_0 - t_1 right)left( underbrace t_0left( ^2 + 1 right)_2 - underbrace t_1left( z_1 right^2 + 1 right)_2 right)\
&= 0
endalign*
So, $rho(z_0,z_1)=d(z_0,z_1)$ and the proof is complete!

Some cool facts about this question:

• The distance $d$ is known as the textbfchordal metric and is interpreted as the euclidean distance between the stereographic projections of two complex numbers. That's why it's called chordal metric (the distance between two points in an sphere is the length of the chord that has the points as its extremes).


• The metric $d$ can be extended to metric in $mathbbC^*$ by setting
  $$dleft( z,infty right) = frac2sqrt 1 + left^2 $$


• $d$ (and its extention) is bounded: the distance between two points is always equal or less than 2.

You can find more about this metric (but not the proof that it's actually a matric) in the books Introduction to Complex Analysis by B.V. Shabat (pages 6 and 7) and Complex Analysis by T.W. Gamelin (section I.3 and especially exercise 6)

share|cite|improve this answer

edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
  – Martin R
  Jun 26 '15 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, you're totally right
  – Charly
  Jul 1 '15 at 15:58
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

This migth be helpful: I'll proof that the function:
$$d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $$ is a metric in the complex numbers. My idea to prove it is to use the stereographic projection of $mathbbC$ into $S^2$ (the unit sphere of $mathbbR^3$) and define the metric $d$ via the euclidean metric in $mathbbR^3$. The technical details are the following:

Let $z=a+ib$ a complex number, then, the stereographic projection is a function $H:mathbbCtoS^2subsetmathbbR^3$ defined by:
$$H(z)=Hleft( a + ib right) = left( ta,tb,1 - t right)quad,quad;t = frac2left^2 + 1$$
The stereographic projection is an inyective function

For any two complex numbers $z_0=a_0+ib_0$ and $z_1=a_1+ib_1$ define a function $rho$ in the following way:
$$rho left( z_0,z_1 right) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right)$$
Is easy to verify that $rho$ is a metric in $mathbbC$:

• $rho (z_0,z_1) = 0 Leftrightarrow d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = 0 Leftrightarrow Hleft( z_0 right) = Hleft( z_1 right) Leftrightarrow z_0 = z_1$


• $rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = d_mathbbR^3left( Hleft( z_1 right),Hleft( z_0 right) right) = rho (z_1,z_0)$


• 
  

  Let $z_2$ be another complex number, then,
  $$rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) le d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right)$$
  (the inequality is given by the triangular inequality of $d_mathbbR^3$).

  
  
  

  By the definition of $rho$ we have $d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right) = rho (z_0,z_1) + rho (z_1,z_2)$, so $$rho (z_0,z_1) le rho (z_0,z_1) + rho (z_1,z_2).$$

  
  

Now, the ``only'' thing left to prove is that $rho (z_0,z_1) = d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $. This demonstration is not very interesnting (it's just a lot of algebraic manipulations). Here goes the details:
beginalign*
rho (z_0,z_1) &= sqrt left( t_0a_0 - t_1a_1 right)^2 + left( t_0b_0 - t_1b_1 right)^2 + left( t_1 - t_0 right)^2nonumber\
&= sqrt left( t_0a_0 right)^2 - 2t_0t_1a_0a_1 + left( t_1a_1 right)^2 + left( t_0b_0 right)^2 - 2t_0t_1b_0b_1 + left( t_1b_1 right)^2 + t_1^2 - 2t_0t_1 + t_0^2nonumber\
&= sqrt t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)
endalign*
Let $A = t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right)$, $B = t_0^2left( a_0^2 + b_0^2 + 1 right)$ and $C = t_1^2left( a_1^2 + b_1^2 + 1 right)$, so $$rho left( z_0,z_1 right) = sqrt A + B + C$$

Now, note that
beginalign*
frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 &= left| z_0 - z_1 right|sqrt frac21 + ^2 sqrt frac21 + z_1 right^2 = sqrt left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 sqrt t_0t_1nonumber\
&= sqrt t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)
endalign*
Set $D=t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)$, so $$d(z_0,z_1)=sqrtD$$
Now,
beginalign*
A &= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) nonumber\
&= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 + a_o^2 - a_0^2 + a_1^2 - a_1^2 + b_o^2 - b_0^2 + b_1^2 - b_1^2 right)nonumber\
&= t_0t_1left( a_o^2 - 2a_0a_1 + a_1^2 + b_o^2 - 2b_ob_1 + b_1^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 right) - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)nonumber\
&= D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)
endalign*
Putting this in $rho left( z_0,z_1 right) = sqrt A + B + C$ we have
$$rho left( z_0,z_1 right) = sqrt D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C$$
so, we just need to proof that $E= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C $ equals zero.

Now,
beginalign*
E &= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)nonumber\
&= a_0^2left( - t_0t_1 + t_0^2 right) + a_1^2left( - t_0t_1 + t_1^2 right) + b_0^2left( - t_0t_1 + t_0^2 right) + b_1^2left( - t_0t_1 + t_1^2 right) + left( t_0 - t_1 right)^2nonumber\
&= a_0^2t_0left( - t_1 + t_0 right) + a_1^2t_1left( - t_0 + t_1 right) + b_0^2t_0left( - t_1 + t_0 right) + b_1^2t_1left( - t_0 + t_1 right) + left( t_0 - t_1 right)^2\
&= left( t_0 - t_1 right)left( a_0^2t_0 - a_1^2t_1 + b_0^2t_0 - b_1^2t_1 + t_0 - t_1 right)\
&= left( t_0 - t_1 right)Bigl( t_0left( a_0^2 + b_0^2 right) - t_1left( a_1^2 + b_1^2 right) + t_0 - t_1 Bigr) \
&= left( t_0 - t_1 right)left( t_0^2 - t_1 z_1 right^2 + t_0 - t_1 right) \
&= left( t_0 - t_1 right)left( underbrace t_0left( ^2 + 1 right)_2 - underbrace t_1left( z_1 right^2 + 1 right)_2 right)\
&= 0
endalign*
So, $rho(z_0,z_1)=d(z_0,z_1)$ and the proof is complete!

Some cool facts about this question:

• The distance $d$ is known as the textbfchordal metric and is interpreted as the euclidean distance between the stereographic projections of two complex numbers. That's why it's called chordal metric (the distance between two points in an sphere is the length of the chord that has the points as its extremes).


• The metric $d$ can be extended to metric in $mathbbC^*$ by setting
  $$dleft( z,infty right) = frac2sqrt 1 + left^2 $$


• $d$ (and its extention) is bounded: the distance between two points is always equal or less than 2.

You can find more about this metric (but not the proof that it's actually a matric) in the books Introduction to Complex Analysis by B.V. Shabat (pages 6 and 7) and Complex Analysis by T.W. Gamelin (section I.3 and especially exercise 6)

share|cite|improve this answer

edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

This migth be helpful: I'll proof that the function:
$$d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $$ is a metric in the complex numbers. My idea to prove it is to use the stereographic projection of $mathbbC$ into $S^2$ (the unit sphere of $mathbbR^3$) and define the metric $d$ via the euclidean metric in $mathbbR^3$. The technical details are the following:

Let $z=a+ib$ a complex number, then, the stereographic projection is a function $H:mathbbCtoS^2subsetmathbbR^3$ defined by:
$$H(z)=Hleft( a + ib right) = left( ta,tb,1 - t right)quad,quad;t = frac2left^2 + 1$$
The stereographic projection is an inyective function

For any two complex numbers $z_0=a_0+ib_0$ and $z_1=a_1+ib_1$ define a function $rho$ in the following way:
$$rho left( z_0,z_1 right) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right)$$
Is easy to verify that $rho$ is a metric in $mathbbC$:

• $rho (z_0,z_1) = 0 Leftrightarrow d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = 0 Leftrightarrow Hleft( z_0 right) = Hleft( z_1 right) Leftrightarrow z_0 = z_1$


• $rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) = d_mathbbR^3left( Hleft( z_1 right),Hleft( z_0 right) right) = rho (z_1,z_0)$


• 
  

  Let $z_2$ be another complex number, then,
  $$rho (z_0,z_1) = d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) le d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right)$$
  (the inequality is given by the triangular inequality of $d_mathbbR^3$).

  
  
  

  By the definition of $rho$ we have $d_mathbbR^3left( Hleft( z_0 right),Hleft( z_1 right) right) + d_mathbbR^3left( Hleft( z_1 right),Hleft( z_2 right) right) = rho (z_0,z_1) + rho (z_1,z_2)$, so $$rho (z_0,z_1) le rho (z_0,z_1) + rho (z_1,z_2).$$

  
  

Now, the ``only'' thing left to prove is that $rho (z_0,z_1) = d(z_0,z_1) = frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 $. This demonstration is not very interesnting (it's just a lot of algebraic manipulations). Here goes the details:
beginalign*
rho (z_0,z_1) &= sqrt left( t_0a_0 - t_1a_1 right)^2 + left( t_0b_0 - t_1b_1 right)^2 + left( t_1 - t_0 right)^2nonumber\
&= sqrt left( t_0a_0 right)^2 - 2t_0t_1a_0a_1 + left( t_1a_1 right)^2 + left( t_0b_0 right)^2 - 2t_0t_1b_0b_1 + left( t_1b_1 right)^2 + t_1^2 - 2t_0t_1 + t_0^2nonumber\
&= sqrt t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)
endalign*
Let $A = t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right)$, $B = t_0^2left( a_0^2 + b_0^2 + 1 right)$ and $C = t_1^2left( a_1^2 + b_1^2 + 1 right)$, so $$rho left( z_0,z_1 right) = sqrt A + B + C$$

Now, note that
beginalign*
frac2leftsqrt 1 + ^2 sqrt 1 + z_1 right^2 &= left| z_0 - z_1 right|sqrt frac21 + ^2 sqrt frac21 + z_1 right^2 = sqrt left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 sqrt t_0t_1nonumber\
&= sqrt t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)
endalign*
Set $D=t_ot_1left( left( a_0 - a_1 right)^2 + left( b_0 - b_1 right)^2 right)$, so $$d(z_0,z_1)=sqrtD$$
Now,
beginalign*
A &= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 right) nonumber\
&= t_0t_1left( - 2a_0a_1 - 2b_ob_1 - 2 + a_o^2 - a_0^2 + a_1^2 - a_1^2 + b_o^2 - b_0^2 + b_1^2 - b_1^2 right)nonumber\
&= t_0t_1left( a_o^2 - 2a_0a_1 + a_1^2 + b_o^2 - 2b_ob_1 + b_1^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 - 2 - a_0^2 - a_1^2 - b_0^2 - b_1^2 right)nonumber\
&= t_0t_1left( left( a_o - a_1 right)^2 + left( b_o - b_1 right)^2 right) - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)nonumber\
&= D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right)
endalign*
Putting this in $rho left( z_0,z_1 right) = sqrt A + B + C$ we have
$$rho left( z_0,z_1 right) = sqrt D - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C$$
so, we just need to proof that $E= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + B + C $ equals zero.

Now,
beginalign*
E &= - t_0t_1left( 2 + a_0^2 + a_1^2 + b_0^2 + b_1^2 right) + t_0^2left( a_0^2 + b_0^2 + 1 right) + t_1^2left( a_1^2 + b_1^2 + 1 right)nonumber\
&= a_0^2left( - t_0t_1 + t_0^2 right) + a_1^2left( - t_0t_1 + t_1^2 right) + b_0^2left( - t_0t_1 + t_0^2 right) + b_1^2left( - t_0t_1 + t_1^2 right) + left( t_0 - t_1 right)^2nonumber\
&= a_0^2t_0left( - t_1 + t_0 right) + a_1^2t_1left( - t_0 + t_1 right) + b_0^2t_0left( - t_1 + t_0 right) + b_1^2t_1left( - t_0 + t_1 right) + left( t_0 - t_1 right)^2\
&= left( t_0 - t_1 right)left( a_0^2t_0 - a_1^2t_1 + b_0^2t_0 - b_1^2t_1 + t_0 - t_1 right)\
&= left( t_0 - t_1 right)Bigl( t_0left( a_0^2 + b_0^2 right) - t_1left( a_1^2 + b_1^2 right) + t_0 - t_1 Bigr) \
&= left( t_0 - t_1 right)left( t_0^2 - t_1 z_1 right^2 + t_0 - t_1 right) \
&= left( t_0 - t_1 right)left( underbrace t_0left( ^2 + 1 right)_2 - underbrace t_1left( z_1 right^2 + 1 right)_2 right)\
&= 0
endalign*
So, $rho(z_0,z_1)=d(z_0,z_1)$ and the proof is complete!

Some cool facts about this question:

• The distance $d$ is known as the textbfchordal metric and is interpreted as the euclidean distance between the stereographic projections of two complex numbers. That's why it's called chordal metric (the distance between two points in an sphere is the length of the chord that has the points as its extremes).


• The metric $d$ can be extended to metric in $mathbbC^*$ by setting
  $$dleft( z,infty right) = frac2sqrt 1 + left^2 $$


• $d$ (and its extention) is bounded: the distance between two points is always equal or less than 2.

You can find more about this metric (but not the proof that it's actually a matric) in the books Introduction to Complex Analysis by B.V. Shabat (pages 6 and 7) and Complex Analysis by T.W. Gamelin (section I.3 and especially exercise 6)

share|cite|improve this answer

edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

share|cite|improve this answer

share|cite|improve this answer

edited Jul 1 '15 at 16:26

edited Jul 1 '15 at 16:26

edited Jul 1 '15 at 16:26

answered Jun 25 '15 at 16:53

Charly

876

answered Jun 25 '15 at 16:53

Charly

876

answered Jun 25 '15 at 16:53

Charly

876

876

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
  – Martin R
  Jun 26 '15 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, you're totally right
  – Charly
  Jul 1 '15 at 15:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
  – Martin R
  Jun 26 '15 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, you're totally right
  – Charly
  Jul 1 '15 at 15:58
  

  
  
  
  

Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
– Martin R
Jun 26 '15 at 17:55

Your metric $rho$ (which is the well-known chordal metric in the extended complex plane) is not the the distance function $d$ defined in the question. Note that in the definition of $d$, the square roots in the denominator are added, not multiplied. So unless I am overlooking something, your calculations are correct, but unrelated to this question.
– Martin R
Jun 26 '15 at 17:55

Oh, you're totally right
– Charly
Jul 1 '15 at 15:58

Oh, you're totally right
– Charly
Jul 1 '15 at 15:58

add a comment | 

up vote
0
down vote

I did a small step before I stuck, but it may be useful for others. So, it seems the following.

The cases $|y|lemin,$ and $|y|gemax,$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^-1$, $v=y^-1$, and $w=z^-1$).

So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics.

Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=sqrt ^2 + sqrt 1 + $, and $v_2=sqrt ^2 + sqrt 1 + $ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds
$$fraccosangle oyx-cosangle zyo=frac v_1v_2.$$

But $$cosangle oyx=frac(y-x,y),$$
$$- cosangle zyo=frac(y-z,y)y.$$

So $$frac(y-x,y)=
frac(y-z,y)left(sqrt y + sqrt ^2right).$$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

add a comment | 

up vote
0
down vote

I did a small step before I stuck, but it may be useful for others. So, it seems the following.

The cases $|y|lemin,$ and $|y|gemax,$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^-1$, $v=y^-1$, and $w=z^-1$).

So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics.

Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=sqrt ^2 + sqrt 1 + $, and $v_2=sqrt ^2 + sqrt 1 + $ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds
$$fraccosangle oyx-cosangle zyo=frac v_1v_2.$$

But $$cosangle oyx=frac(y-x,y),$$
$$- cosangle zyo=frac(y-z,y)y.$$

So $$frac(y-x,y)=
frac(y-z,y)left(sqrt y + sqrt ^2right).$$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

add a comment | 

up vote
0
down vote

up vote
0
down vote

I did a small step before I stuck, but it may be useful for others. So, it seems the following.

The cases $|y|lemin,$ and $|y|gemax,$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^-1$, $v=y^-1$, and $w=z^-1$).

So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics.

Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=sqrt ^2 + sqrt 1 + $, and $v_2=sqrt ^2 + sqrt 1 + $ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds
$$fraccosangle oyx-cosangle zyo=frac v_1v_2.$$

But $$cosangle oyx=frac(y-x,y),$$
$$- cosangle zyo=frac(y-z,y)y.$$

So $$frac(y-x,y)=
frac(y-z,y)left(sqrt y + sqrt ^2right).$$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

I did a small step before I stuck, but it may be useful for others. So, it seems the following.

The cases $|y|lemin,$ and $|y|gemax,$ are simple and already known to us (see Arian’s answer and here; the second case is reduced to the first by substitution $u=x^-1$, $v=y^-1$, and $w=z^-1$).

So it suffices to consider a case $| x | < | y | < | z |.$ Fix $x$, $z$ and $|y|$. Answering a question for which value of $y$ the left part of the inequality attains minimum, I use geometric optics.

Let there be light. Let the speed of light in the disk with radius $|y|$ centered at the origin $o$ be $v_1=sqrt ^2 + sqrt 1 + $, and $v_2=sqrt ^2 + sqrt 1 + $ outside the disk. Fermat's principle states that the light travels the path which takes the least time. From Fermat principle may be derived known Snell–Descartes law of refraction, which yeilds
$$fraccosangle oyx-cosangle zyo=frac v_1v_2.$$

But $$cosangle oyx=frac(y-x,y),$$
$$- cosangle zyo=frac(y-z,y)y.$$

So $$frac(y-x,y)=
frac(y-z,y)left(sqrt y + sqrt ^2right).$$

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

share|cite|improve this answer

share|cite|improve this answer

edited Apr 13 '17 at 12:21

Community♦

1

edited Apr 13 '17 at 12:21

Community♦

1

edited Apr 13 '17 at 12:21

Community♦

1

1

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

answered Jan 23 '15 at 13:55

Alex Ravsky

36.6k32074

36.6k32074

add a comment | 

add a comment | 

 

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