Complex numbers: $|\frac{1}{x}-\frac{1}{y}| = \frac{|x-y|}{|x||y|}$? I need the result for a proof, but I can't seem to verify it. If $x,y$ are distinct nonzero complex numbers, why is it true that
$|\frac{1}{x}-\frac{1}{y}| = \frac{|x-y|}{|x||y|}$? 
Starting with the left hand side, I've gotten to the term 
$|\frac{\bar{x}|y|^{2}-\bar{y}|x|^2}{|x|^{2}|y|^{2}}|$
but algebraically can't seem to equate the sides.
(Also, TeX tips/tricks are very much welcomed for ways of writing the modulus | | bars)
 A: $\vert\frac{1}{x}-\frac{1}{y}\vert = \vert \frac{y}{yx}-\frac{x}{xy} \vert = \vert \frac{y-x}{xy} \vert = \frac{\vert{y-x}\vert}{\vert xy \vert}$. I am assuming you're talking about the usual norm, which is compatible with multiplication, and division (by non zero numbers).
A: $\begin{align}
\left\lvert \frac 1 x - \frac 1 y\right\rvert 
& = \left\lvert \frac y{xy}-\frac x{xy}\right\rvert
\\ & = \left\lvert \frac {y-x}{xy}\right\rvert
\\ & = \frac {\lvert y-x\rvert}{\lvert xy\rvert}
\\ & = \frac {\lvert y-x\rvert}{\lvert x\rvert\lvert y\rvert}
\end{align}$
A: *

*Why is $\dfrac 1 x - \dfrac 1 y = \dfrac{y-x}{xy}$?

*Why is $|ab|=|a||b|$?

*Why is $|a/b|=|a|/|b|$?


The first bullet point above gets you from $\displaystyle\left|\frac 1 x - \frac 1 y\right|$ to $\displaystyle\left|\frac{x-y}{xy}\right|$.  The third bullet point gets you from there to $\displaystyle\frac{|x-y|}{|xy|}$.  The second bullet point gets you from there to $\displaystyle\frac{|x-y|}{|x||y|}$.
The answer to the first question comes from the fact that the complex numbers form a field, i.e. a system in which multiplication and addition satisfy certain basic laws you first learned for real numbers: They are both commutatitive and associative; both have identity elements, which differ from each other; everything has an additive inverse; everything except the additive identity has a multiplicative inverse; and multiplication distributes over addition.  And after you've established all that, you need to do a bit of algebra with fractions.
To answer the second bullet point, suppose $a=p+iq$, $b=r+is$, and $p,q,r,s$ are real, then $|a|=\sqrt{p^2+q^2}$ and similarly for $b$, and $ab= (pr - qs) + i(ps + rq)$, so $|ab|=\sqrt{(pr-qs)^2+(ps+rq)^2}$.  So you need to check that
$$
(p^2+q^2)(r^2+s^2) = (pr-qs)^2+(ps+rq)^2.
$$
A similar thing applies to division.
