# complex numbers

• May 13th 2009, 04:38 AM
stud_02
complex numbers
$P (Z_1)$ and $Q (Z_2)$ are complex numbers in argand diagram and $O$ is the origin.
show $OP$ and $OQ$ are perpendicular if

$|Z_1 - Z_2| = |Z_1 + Z_2|$
• May 13th 2009, 08:26 AM
Soroban
Hello, stud_02!

Quote:

$P (Z_1)$ and $Q (Z_2)$ are complex numbers in argand diagram and $O$ is the origin.

Show $OP$ and $OQ$ are perpendicular if: . $|Z_1 - Z_2| = |Z_1 + Z_2|$

Code:

          Q * - - - - - - - - - - - * R           /  *                *  /           /    *          *    /         /        *    *        /     Z2 /          *          /       /        *    *        /       /    *          *    /     /  *                *  /   O * - - - - - - - - - - - * P               Z1

$\overrightarrow{Z_1} \:=\:\overrightarrow{OP}\:\text{ and }\:\overrightarrow{Z_2} \:=\:\overrightarrow{OQ}$

Then: . $\begin{array}{ccc}\overrightarrow{Z_1}+\overrighta rrow{Z_2} &=& \overrightarrow{OR} \\ \\[-4mm] \overrightarrow{Z_1}-\overrightarrow{Z_2} &=& \overrightarrow{QP} \end{array}
$

Hence . $\begin{array}{ccc}|\overrightarrow{Z_1} + \overrightarrow{Z_2}|&=& OR \\ \\[-4mm] |\overrightarrow{Z_1}-\overrightarrow{Z_1}| &=& QP \end{array}$

If $|\overrightarrow{Z_1} - \overrightarrow{Z_2}| \:=\:|\overrightarrow{Z_1}+\overrightarrow{Z_2}|$, then: . $QP \:=\:OR$

. . And we have a parallelogram with equal diagonals.

A parallelogram with equal diagonals is a rectangle.

. . Therefore: . $OP \perp OQ$

• May 13th 2009, 05:47 PM
Soroban
Hello again, stud_02!

An algebraic proof is probably expected . . .

Quote:

$P (Z_1)$ and $Q (Z_2)$ are complex numbers in the argand diagram and $O$ is the origin.

Show $OP \perp OQ$ if $|Z_1-Z_2| \:=\: |Z_1+Z_2|$

Let: . $\begin{array}{ccc}Z_1 &=& a + bi \\ Z_2 &=& c + di\end{array}$

We know that: . $Z_1 \perp Z_2\:\text{ if }\:Z_1\!\cdot\! Z_2 \:=\: 0$

. . That is: . $(a+bi)\cdot(c+di) \:=\:0 \quad\Rightarrow\quad ac + bd \:=\:0$ .[1]

We will show that [1] is true.

$Z_1 - Z_2 \;=\;(a-c) + (b-d)i$

. . $|Z_1-Z_2| \;=\;\sqrt{(a-c)^2 + (b-d)^2}$

$Z_1+Z_2 \;=\;(a+c)+(b+d)i$

. . $|Z_1+Z_2| \;=\;\sqrt{(a+c)^2 + (b+d)i}$

Since $|Z_1-Z_2| \:=\:|Z_1+Z_2|$, we have: . $\sqrt{(a-c)^2 + (b-d)^2} \;=\;\sqrt{(a+c)^2+(b+d)^2}$

Square both sides: . $(a-c)^2 + (b-d)^2 \;=\;(a+c)^2 + (b+d)^2$

Expand: . $a^2-2ac + c^2 + b^2 - 2bd + d^2 \;=\;a^2 + 2ac + c^2 + b^2 + 2bd + d^2$

. . which simplifies to: . $4ac + 4bd \:=\:0 \quad\Rightarrow\quad\boxed{ ac + bd \:=\:0}$ . Q.E.D.

• May 13th 2009, 06:29 PM
pankaj
Quote:

Originally Posted by stud_02
$P (Z_1)$ and $Q (Z_2)$ are complex numbers in argand diagram and $O$ is the origin.
show $OP$ and $OQ$ are perpendicular if

$|Z_1 - Z_2| = |Z_1 + Z_2|$

$
|z_{1}-z_{2}|^2=|z_{1}+z_{2}|^2
$

$
|z_{1}|^2+|z_{2}|^2-z_{1} \bar z_{2}-\bar z_{1} z_{2}=
|z_{1}|^2+|z_{2}|^2+z_{1}\bar z_{2}+\bar z_{1} z_{2}
$

$
2(z_{1}\bar z_{2}+\bar z_{1} z_{2})=0
$

$z_{1}\bar z_{2}=-\bar z_{1} z_{2}$

$
\frac{z_{1}}{z_{2}}+\frac{\bar z_{1}}{\bar z_{2}}=0
$

$\frac{z_{1}}{z_{2}}$ is purely imaginary

$
arg(\frac{z_{1}}{z_{2}})=\frac{\pi}{2}
$

$arg(z_{1})-arg(z_{2})=\frac{\pi}{2}$

$arg(\overrightarrow{OP})-arg(\overrightarrow{OQ})=\frac{\pi}{2}$

Hence, $\overrightarrow{OP}$ is perpendicular to $\overrightarrow{OQ}$