Can anyone help me show this equation?
For a complex z != 0
Thanks
$\displaystyle \left | \bar{z} - \frac{1}{z} \right | = \left | |z| - \frac{1}{|z|} \right |$
Let z = a + bi. Then $\displaystyle \bar{z} = a - bi$ and $\displaystyle |z| = | \bar{z} | = \sqrt{a^2 + b^2}$. So the LHS becomes:
$\displaystyle \left | \bar{z} - \frac{1}{z} \right | = \left | a - bi - \frac{1}{a + bi} \right | = \left |\frac{(a - bi)(a^2 + b^2) - (a - bi)}{a^2 + b^2} \right |$
= $\displaystyle \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}$
Now for the RHS:
$\displaystyle \left | |z| - \frac{1}{|z|} \right | = \left | \sqrt{a^2 + b^2} - \frac{1}{\sqrt{a^2 + b^2}} \right |$
Since this is real we don't need the outside | | bars.
= $\displaystyle \frac{(a^2 + b^2) - 1}{\sqrt{a^2 + b^2}}$
which is the same as the LHS.
-Dan
$\displaystyle \left|\overline{z}-\frac{1}{z}\right|\ =\ \left||z|-\frac{1}{|z|}\right|\ \Leftrightarrow\ /z \neq 0/\ \Leftrightarrow$
$\displaystyle \left|\overline{z}-\frac{1}{z}\right|\cdot|z|\ =\ \left||z|-\frac{1}{|z|}\right|\cdot|z|\ \Leftrightarrow$
$\displaystyle \left|\overline{z}\cdot z-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow$
$\displaystyle \left||z|^2-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow$
$\displaystyle 1\ =\ 1$, which is true for all $\displaystyle z$
$\displaystyle z \not = 0$, else $\displaystyle z$ can be anything.
-Kristofer.
I'll do it anyway, in case someone wants it. Because I'm just that nice a guy!
$\displaystyle \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | $
Note that the first factor (the fraction) is real. So we can bring it outside the | |:
$\displaystyle \left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot |a - bi|$
= $\displaystyle \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot \sqrt{a^2 + b^2}$
= $\displaystyle \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}$
-Dan
Here is another way.
$\displaystyle z \not= 0\quad \Rightarrow \quad \frac{1}{z} = \frac{{\bar z}}{{\left| z \right|^2 }}\quad \& \quad \left| z \right| = \left| {\bar z} \right|$
$\displaystyle \begin{array}{rcl}
\left| {\bar z - \frac{1}{z}} \right| & = & \left| {\bar z - \frac{{\bar z}}{{\left| z \right|^2 }}} \right| \\
& = & \frac{{\left| {\bar z} \right|}}{{\left| z \right|}}\left| {\left| z \right| - \frac{1}{{\left| z \right|}}} \right| \\
& = & \left| {\left| z \right| - \frac{1}{{\left| z \right|}}} \right| \\
\end{array}
$
Unless I'm missing something (always possible!) I don't see how that condition would affect anything.
-Dan
Ah! Gotcha. Then you wouldn't be able to remove the | | bars. (I hadn't thought of that.) But since they are there on both sides of the equality we are trying to show both sides would still be equal; you would simply add a negative sign to both the LHS and the RHS to make it correct.
-Dan