My Math Forum area of quadrilateral

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 December 17th, 2011, 07:45 PM #1 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 area of quadrilateral the area of quadrilateral formed by $z\;,\bar{z}\;,\frac{1}{z}$ and $\frac{1}{\bar{z}}$
December 18th, 2011, 08:34 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Hello, panky!

Quote:
 $\text{The area of quadrilateral formed by: }\:z,\;\bar{z},\;\frac{1}{z},\,\text{ and }\,\frac{1}{\bar{z}}$

$\text{Let }\,z \:=\:a\,+\,bi \;\;\;\Rightarrow\;\;\;\bar{z} \:=\:a\,-\,bi$

$\text{Then: }\:\begin{Bmatrix}\frac{1}{z}=&\frac{1}{a\,+\,bi}=&\frac{a\,-\,bi}{a^2\,+\,b^2} \\ \\ \\ \frac{1}{\bar{z}}=&\frac{1}{a\,-\,bi}=&\frac{a\,+\,bi}{a^2\,+\,b^2} \end{Bmatrix}=$

Plot the four points.
The graph looks like this . . .

Code:
        |         * A
|       * |
|   D *   |
|   *     |
|   |     |
- - + - * - - * - - -
|  P|     |Q
|   *     |
|   C *   |
|       * |
|         * B
$\text{We have six points: }\:\begin{bmatrix} A\,(a,\,b) \\ \\ \\ B(a,\,-b) \\ \\ \\ C\left(\frac{a}{a^2+b^2},\,\frac{-b}{a^2+b^2}\right) \\ \\ \\ D\left(\frac{a}{a^2+b^2},\,\frac{b}{a^2+b^2}\right ) \\ \\ \\ \\ P(a,\,0) \\ \\ \\ \\ Q\left(\frac{a}{a^2+b^2},\,0\right) \end{bmatrix}$

$\text{The area of a trapezoid is: }\:\frac{h}{2}(b_1\,+\,b_2)$

$h \:=\:PQ \:=\:a\,-\,\frac{a}{a^2+b^2} \:=\:a\,\left(\frac{a^2+b^2-1}{a^2+b^2}\right)$

$b_1 \:=\:AB \:=\:2b$

$b_2 \:=\:\text{{C}D} \:=\:\frac{2b}{a^2+b^2}$
[color=beige]. . [/color]$b_1\,+\,b_2 \:=\:2b\,+\,\frac{2b}{a^2+b^2} \:=\:2b\left(1 \,+\,\frac{1}{a^2+b^2}\right) \:=\:2b\,\left(\frac{a^2+b^2+1}{a^2+b^2}\right)$

$\text{Area} \;=\;\frac{1}{2}\,\cdot\,a\,\left(\frac{a^2+b^2-1}{a^2+b^2}\right)\,\cdot\,2b\,\left(\frac{a^2+b^2 +1}{a^2+b^2}\right)$

[color=beige]. . . . .[/color]$=\;ab\,\left(\frac{a^2+b^2-1}{a^2+b^2}\right)\left(\frac{a^2+b^2-1}{a^2+b^2}\right)$

[color=beige]. . . . .[/color]$=\; ab\,\left(\frac{(a^2+b^2)^2\,-\,1}{(a^2+b^2)^2}\right)$

$\text{Since }\,a^2\,+\,b^2 \,=\,|z|^2,\,\text{ we have:}$

[color=beige]. . [/color]$\text{Area} \;=\;ab\left(\frac{|z|^4\,-\,1}{|z|^4}\right)$

 December 18th, 2011, 08:39 PM #3 Senior Member   Joined: Jul 2011 Posts: 400 Thanks: 15 Re: area of quadrilateral thank soroban

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