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Math Help - complex numbers

  1. #1
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    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|
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  2. #2
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    Hello, stud_02!

    How about a geometric proof?


    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}<br />


    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



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  3. #3
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    Hello again, stud_02!

    An algebraic proof is probably expected . . .


    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.

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  4. #4
    Senior Member pankaj's Avatar
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    Quote Originally Posted by stud_02 View Post
    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|

     <br />
|z_{1}-z_{2}|^2=|z_{1}+z_{2}|^2<br />

     <br />
|z_{1}|^2+|z_{2}|^2-z_{1} \bar z_{2}-\bar z_{1} z_{2}=<br />
|z_{1}|^2+|z_{2}|^2+z_{1}\bar z_{2}+\bar z_{1} z_{2}<br />

     <br />
2(z_{1}\bar z_{2}+\bar z_{1} z_{2})=0<br />

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

     <br />
\frac{z_{1}}{z_{2}}+\frac{\bar z_{1}}{\bar z_{2}}=0<br />

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

     <br />
arg(\frac{z_{1}}{z_{2}})=\frac{\pi}{2}<br />

    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}
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