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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)

Michael Hardy
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zha
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    use \left| to make a | on the left side, and let Tex decide how tall they should be. ALWAYS end it with a \right|. This trick works for just about any delimiter. – Zach Stone Jun 11 '15 at 00:15
  • Also \lvert and \rvert produces neater vertical bars than just the | symbol; i.e., this $\lvert x - y \rvert$ vs $|x - y|$. Can be paired with \left and \right, as well as the bonus \lVert and \rVert for the double bars :) – pjs36 Jun 11 '15 at 00:42

3 Answers3

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$\vert\frac{1}{x}-\frac{1}{y}\vert = \vert \frac{y}{yx}-\frac{x}{xy} \vert = \vert \frac{y-x}{xy} \vert = \frac{\vert{y-x}\vert}{\vert xy \vert}$. I am assuming you're talking about the usual norm, which is compatible with multiplication, and division (by non zero numbers).

mich95
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$\begin{align} \left\lvert \frac 1 x - \frac 1 y\right\rvert & = \left\lvert \frac y{xy}-\frac x{xy}\right\rvert \\ & = \left\lvert \frac {y-x}{xy}\right\rvert \\ & = \frac {\lvert y-x\rvert}{\lvert xy\rvert} \\ & = \frac {\lvert y-x\rvert}{\lvert x\rvert\lvert y\rvert} \end{align}$

Graham Kemp
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  • 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.

Michael Hardy
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