# To verify d defines a metric on D

Let $D := \{ z \in \mathbb{C} : \left | z \right | \le 1 \}$ be a unit disk in the complex plane and define $$d(z,w) := \begin{cases} \left| z-w \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \\ \end{cases}$$

Verify that $d$ defines a metric on $D$.

SOLUTION: Checking the metric axioms

$M_1: d(z,w) \ge 0$:

$d(z,w) \ge 0$ holds from the definition.

$M_2:d(z,w)=0 \Leftrightarrow z=w$:

$$d(z,w)=0= \begin{cases} \left| z-w \right| & \text{ if } \arg (z)=\arg(w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \end{cases}$$

We get $z=w$ on the other hand if $z=w$, $$d(z,w)= \begin{cases} \left| z-z \right| & \text{ if } \arg (z)=\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero} \\ \left| z \right|+\left| z \right| & \text{ otherwise } \\ \end{cases}$$

$$|z-z|=0 \\ \text{so } d(z,w)=0 \Leftrightarrow z=w \text{ holds} \\$$

${{M}_{3}};\,d(z,w)=d(w,z)$

$$d\left( z,w \right)=\begin{cases} \left| z-w \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| z \right|+\left| w \right| & \text{ otherwise } \\ \end{cases}$$ $$=d\left( w,z \right)=\begin{cases} \left| w-z \right| & \text{ if } \arg (z) =\arg (w) \text{ or if one of } z \text{ and } w \text{ is zero } \\ \left| w \right|+\left| z \right| & \text{ otherwise } \\ \end{cases}$$ $$\text{so }d(z,w)=d(w,z) \text{ holds}$$

${{M}_{4}};\text{ Triangle Inequality}: d(z,w)\le d(z,x)+d\left( x,w \right)$

Somebody to assist me in axioms $M_2$ , and $M_4$. Regards.

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It looks like you just cut and pasted from a LaTeX file... In general that doesn't work. – Thomas Apr 14 '12 at 15:23
This needs $\LaTeX$ help. Please review-I find it unreadable as is. The box around some of the start is usually a mismatched delimeter, like a brace. – Ross Millikan Apr 14 '12 at 15:24
Even if it compiled on StackExchange, this is still an abomination. So many \,s. – Chris Eagle Apr 14 '12 at 15:39
As a pointer, one can use \text{} rather than what you are using. \text{} will allow you to put actual words inside the math. For example, $$These\, are\, words.$$ becomes, with \text{}, $$\text{These are words.}$$ – 000 Apr 14 '12 at 16:17
In France, this distance is sometimes called distance SNCF. This refers to the fact that the railway network is (mainly) organized radially around a center (Paris), hence to go from city A to city B the quickest route is often to go from A to Paris and then from Paris to B (of course there is something of an overstatement in this description). – Did Apr 14 '12 at 19:17
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Dislcaimer: If any of this is erroneously false, inform me rather than downvoting and I will delete this answer.

From what I understand, $M_2$ is satisfied.

You can show that if $z=w$, it is evident that $d(z,w)=0$ via the definition of the function in the case of $\arg(z)=\arg(z)$. (It is clear that $|z-w|=0$ if $z=w$.)

However, proving this is the only such case that $d(z,w)$ is $0$ follows from definition. Simply notice that $|z-w|$ and $|z|+|w|$ are not zero unless $z=w$. (I don't know how this needs further elaboration or if it does?)

The triangle inequality (in this case?) reduces to simply showing: $$d(z,w) \le d(z,x)+d(x,w)$$ There are four cases via the first definition: \begin{align} |z-w|&\le |z-x|+|x-w|\\ |z-w|&\le |z-x|+|x|+|w|\\ |z-w|&\le |z|+|x|+|x-w|\\ |z-w|&\le |z|+|x|+|x|+|w| \end{align}

I'll prove these four one by one (as I am not sure if any of them are absurd cases):

\begin{align} |z-w|&\le |z-x|+|x-w|\\ \text{Let } a&=z-x\\ \text{and } b&=x-w\\ |a+b| &\le |a|+|b|\\ \therefore |z-w|&\le |z-x|+|x-w| \end{align} (i.e. the case reduces to the definition of the triangle inequality with regard to complex numbers)

For case two, $d(z,x)=|z-x|$ and $d(x,w)=|x|+|w|$ tell us that $z=0$. If $z=0$, then: \begin{align} |z-w|&=|-w|=|w|\\ \text{and } |z-x|+|x|+|w|&=|-x|+|x|+|w|\\ &=2|x|+|w|\\ \text{thus } |w|&\le 2|x|+|w|\\ 0 &\le 2|x|\\ \therefore |z-w| &\le |z-x|+|x|+|w| \end{align}

Case three implies that $w=0$, so:

\begin{align} |z| &\le |z|+|x|+|x|\\ 0 &\le 2|x|\\ \therefore |z-w| &\le |z|+|w|+|x-w| \end{align}

Case four:

First, let:

\begin{align} z&=a_z+b_zi\\ w&=a_w+b_wi\\ x&=a_x+b_xi \end{align}

Then, we have:

\begin{align} |z-w|&\le |z|+2|x|+|w|\\ \sqrt{(a_z-a_w)^2+(b_z-b_w)^2}&\le \sqrt{a_z^2+b_z^2}+2\sqrt{a_x^2+b_x^2}+\sqrt{a_w^2+b_w^2}\\ (a_z-a_w)^2+(b_z-b_w)^2 &\le a_z^2+b_z^2+a_w^2+b_w^2+2|x|+4|x||z|+2|z||w|+4|x||w|\\ -2(a_za_w+b_zb_w) &\le 2|x|+4(|x||z|+|x||w|)+2|z||w|\\ 0 & \le |x|+2(|x||z|+|x||w|)+|z||w|+a_za_w+b_zb_w \end{align}

$\arg(z)=\arg(w)$ ( http://www.wolframalpha.com/input/?i=arg%28z%29%3Darg%28w%29 ) is the only such situation for case four, therefore the product of their real parts and their imaginary parts cannot be negative and, hence, the final inequality is true.

All that is left now are the four other similar cases:

\begin{align} |z|+|w|&\le |z-x|+|x-w|\\ |z|+|w|&\le |z-x|+|x|+|w|\\ |z|+|w|&\le |z|+|x|+|x-w|\\ |z|+|w|&\le |z|+|x|+|x|+|w| \end{align}

Case one follows equality (namely, since neither $z$ nor $w$ are $0$, the inequality reduces to an equality because $x$ must be $0$. (The argument of $x$ cannot be equal to both $w$ and $z$. If it were, then the left side would be $|z-w|$ rather than $|z|+|w|$.))

Case four is straight to the point. It reduces to $0\le 2|x|$.

Addendum: I don't know how to prove case two and three. This is my best attempt at an answer and I hope it aids you.

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The problem is much easier to think about if you build a good mental picture of how the metric $d$ works. Two complex numbers with the same argument lie on the same ray from the origin, and the $d$-distance between them is simply the straight-line distance between them along that ray. Two complex numbers $w$ and $z$ with different arguments lie on different rays from the origin; to get $d(w,z)$, you add the straight-line distance from $w$ to $0$ to the straight-line distance from $0$ to $z$.

Think of the complex plane as a ‘hedgehog’: its body is $0$, and its spines are the rays emanating from the origin, one for each possible value of the argument in $[0,2\pi)$. You can travel along a spine, but you can’t jump from one spine to another. If two points are on the same spine, their $d$-distance is just the distance between them along that spine. If they’re on different spines, to get from one to the other you have to go by way of the origin, so you have to add their distances to the origin to get their $d$-distance.

To see how this picture helps, lets look at the two cases that are omitted from Limitless’s answer:

\begin{align*} |z|+|w|&\le |z-x|+|x|+|w|\tag{1}\\ |z|+|w|&\le |z|+|x|+|x-w|\tag{2}\\ \end{align*}

These are really the same case: $w$ and $z$ are on different spines, and $x$ is on the same spine as one of them. I’ll look at $(2)$, with $x$ on the same spine as $w$. To get from $z$ to either $x$ or $w$, you have to go along $z$’s spine to the origin, so in both cases you’re going to get a $|z|$ term. Then either you go straight from $0$ to $w$ along the spine containing $w$ and $x$, or you go from $0$ to $x$ and then to $w$, in both cases along the spine containing $w$ and $x$. Distances along that spine are measured exactly as they are in the usual metric on $[0,\infty)$ in $\mathbb R$. Thus, ignoring the common part $|z|$ of the total distances, $d(z,w)\le d(z,x)+d(x,w)$ just says in this case that in $\mathbb R$ the distance from $0$ to $|w|$ is at most the distance from $0$ to $|x|$ plus the distance from $|x|$ to $|w|$: $$|w|=|0-w|\le|0-x|+|x-w|\le|x|+|x-w|\;.\tag{3}$$ This is just the ordinary triangle inequality for real numbers.

Note that I didn’t really need the geometric picture to get this; it just made it a little easier to find $(3)$, after which of course we can simply add $|z|$ to both sides to get $(2)$. However, the geometric picture can also be used to simplify the rest of the verification of the triangle inequality, and it makes the other properties of a metric completely obvious.

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 Why thank you for that elucidation. I never considered that approach--it seemed that the triangle inequality was somewhere. – 000 Apr 15 '12 at 12:50