A complex equation

**changing_seasons** · December 8th 2006, 12:27 PM

Can anyone help me show this equation?

For a complex z != 0

Thanks

**topsquark** · December 8th 2006, 01:20 PM

Originally Posted by changing_seasons

Can anyone help me show this equation?

For a complex z != 0

Thanks

$\left | \bar{z} - \frac{1}{z} \right | = \left | |z| - \frac{1}{|z|} \right |$

Let z = a + bi. Then $\bar{z} = a - bi$ and $|z| = | \bar{z} | = \sqrt{a^2 + b^2}$ . So the LHS becomes:
$\left | \bar{z} - \frac{1}{z} \right | = \left | a - bi - \frac{1}{a + bi} \right | = \left |\frac{(a - bi)(a^2 + b^2) - (a - bi)}{a^2 + b^2} \right |$

= $\left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}$

Now for the RHS:
$\left | |z| - \frac{1}{|z|} \right | = \left | \sqrt{a^2 + b^2} - \frac{1}{\sqrt{a^2 + b^2}} \right |$

Since this is real we don't need the outside | | bars.

= $\frac{(a^2 + b^2) - 1}{\sqrt{a^2 + b^2}}$

which is the same as the LHS.

-Dan

**TriKri** · December 8th 2006, 01:31 PM

$\left|\overline{z}-\frac{1}{z}\right|\ =\ \left||z|-\frac{1}{|z|}\right|\ \Leftrightarrow\ /z \neq 0/\ \Leftrightarrow$

$\left|\overline{z}-\frac{1}{z}\right|\cdot|z|\ =\ \left||z|-\frac{1}{|z|}\right|\cdot|z|\ \Leftrightarrow$

$\left|\overline{z}\cdot z-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow$

$\left||z|^2-1\right|\ =\ \left||z|^2-1\right|\ \Leftrightarrow$

$1\ =\ 1$ , which is true for all $z$

$z \not = 0$ , else $z$ can be anything.

-Kristofer.

**changing_seasons** · December 8th 2006, 01:53 PM

Originally Posted by topsquark

= $\left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}$

Thanks for the help, but could you please elaborate on this final step?

NEVER MIND I got it!

**topsquark** · December 8th 2006, 02:02 PM

Originally Posted by changing_seasons

Thanks for the help, but could you please elaborate on this final step?

NEVER MIND I got it!

I'll do it anyway, in case someone wants it. Because I'm just that nice a guy!

$\left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right |$

Note that the first factor (the fraction) is real. So we can bring it outside the | |:

$\left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot |a - bi|$

= $\frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot \sqrt{a^2 + b^2}$

= $\frac{a^2 + b^2 - 1}{\sqrt{a^2 + b^2}}$

-Dan

**TriKri** · December 8th 2006, 02:07 PM

Originally Posted by topsquark

Note that the first factor (the fraction) is real. So we can bring it outside the | |:

$\left | \frac{(a^2 + b^2 - 1)}{a^2 + b^2} \cdot (a - bi) \right | = \frac{a^2 + b^2 - 1}{a^2 + b^2} \cdot |a - bi|$

What about if $a^2 + b^2 - 1 < 0$ ?

-Kristofer

**Plato** · December 8th 2006, 02:20 PM

Here is another way.
$z \not= 0\quad \Rightarrow \quad \frac{1}{z} = \frac{{\bar z}}{{\left| z \right|^2 }}\quad \& \quad \left| z \right| = \left| {\bar z} \right|$

$\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} $

**topsquark** · December 8th 2006, 04:38 PM

Originally Posted by TriKri

What about if $a^2 + b^2 - 1 < 0$ ?

-Kristofer

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

Thread: A complex equation

LinkBack

Thread Tools

Display

A complex equation

Similar Math Help Forum Discussions

Complex equation

Complex equation

complex equation

Complex equation

Complex equation

Search Tags