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Thread: complex number

  1. #1
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    complex number

    Given that $\displaystyle z=x+yi$ and $\displaystyle w=\frac{z+8i}{z-6}$, where z is not 6 . If w is totally imaginary , show that $\displaystyle x^2+y^2+2x-48=0$
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  2. #2
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    Complex numbers

    Hello again mathaddict.
    Given that and , where z is not 6 . If w is totally imaginary , show that
    Are you sure you have copied down the question correctly?

    I think the expression should be $\displaystyle w=\frac{z+8}{z-6}$.

    You then:

    • Replace $\displaystyle z$ by $\displaystyle x+iy$ in the top and bottom
    • Collect together reals and imaginaries on the top and on the bottom
    • Multiply top and bottom by $\displaystyle (x-6)+iy$, which is the conjugate of the bottom

    Don't multiply everything out at this stage. Just find the real part of the top. This must be zero if w is wholly imaginary. This will then give you $\displaystyle x^2+y^2+2x-48=0$

    Grandad
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  3. #3
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    Hello, mathaddict!

    Grandad caught the typo . . .



    Given: .$\displaystyle z\:=\:x+yi\:\text{ and }\:w=\frac{z+{\color{red}8}}{z-6}\;\text{ where }z \neq 6.$

    If $\displaystyle w$ is totally imaginary , show that $\displaystyle x^2+y^2+2x-48\:=\:0$

    Substituting $\displaystyle z \:=\:z+yi$, we have: .$\displaystyle w \;=\;\frac{(x+yi) + 8}{(x+yi) - 6} \;=\;\frac{(x+8) + yi}{(x-6) + yi}$


    Multiply top and bottom by $\displaystyle (x-6)-yi$

    $\displaystyle w \;=\;\frac{(x+8)+yi}{(x-6)+yi}\;\cdot\;{\color{blue}\frac{(x-6) - yi}{(x-6) - yi}} \;=\;\frac{(x+8)(x-6) - y(x+8)i + y(x-6)i - y^2i^2}{(x-6)^2+y^2} $

    . . $\displaystyle = \;\frac{x^2+2x-48 + y^2 + [y(x-6) - y(x+8)]i}{(x-6)^2+y^2} $

    . . $\displaystyle = \;\frac{(x^2+y^2+2x-48) - 14yi}{(x-6)^2+y^2}$

    . . $\displaystyle = \;\underbrace{\frac{x^2+y^2+2x-48}{(x-6)^2+y^2}}_{\Re} - \underbrace{\frac{14yi}{(x-6)^2+y^2}i }_{\Im}$


    Since $\displaystyle w$ is purely imaginary, $\displaystyle \Re \,=\,0.$

    Therefore: .$\displaystyle \frac{x^2+y^2+2x-48}{(x-6)^2+y^2} \:=\:0 \quad\Rightarrow\quad x^2 + y^2 +2x-48 \:=\:0$

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    Re :

    Thank you very much again , Soroban and Grandad !!
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