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Math Help - complex number

  1. #1
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    complex number

    Given that z=x+yi and w=\frac{z+8i}{z-6}, where z is not 6 . If w is totally imaginary , show that 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 w=\frac{z+8}{z-6}.

    You then:

    • Replace z by x+iy in the top and bottom
    • Collect together reals and imaginaries on the top and on the bottom
    • Multiply top and bottom by (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 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: . z\:=\:x+yi\:\text{ and }\:w=\frac{z+{\color{red}8}}{z-6}\;\text{ where }z \neq 6.

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

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


    Multiply top and bottom by (x-6)-yi

    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}

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

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

    . . = \;\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 w is purely imaginary, \Re \,=\,0.

    Therefore: . \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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  4. #4
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    Re :

    Thank you very much again , Soroban and Grandad !!
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