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Thread: Complex number question

  1. #1
    Senior Member Stroodle's Avatar
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    Complex number question

    If $\displaystyle z=x+yi$ find the values of $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle \frac{z-1}{z+1}=z+2$

    Thanks for your help!
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \frac{x-1+yi}{x+1+yi}=x+2+yi\Rightarrow x-1+yi=x^2+3x-y^2+2+i(2xy+3y)\Rightarrow$

    $\displaystyle \Rightarrow\left\{\begin{array}{ll}x^2+3x-y^2+2=x-1\\2xy+3y=y\end{array}\right.\Rightarrow\left\{\be gin{array}{ll}x^2-y^2+2x=-3\\2y(x+1)=0\end{array}\right.$

    Now, you have to solve the system. Can you do that?
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  3. #3
    Super Member Gamma's Avatar
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    Quote Originally Posted by Stroodle View Post
    If $\displaystyle z=x+yi$ find the values of $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle \frac{z-1}{z+1}=z+2$

    Thanks for your help!
    $\displaystyle \frac{z-1}{z+1}=z+2\Rightarrow z-1=z^2+3z+2 \Rightarrow 0$$\displaystyle =z^2+2z+3 \Rightarrow z=\frac{-2\pm \sqrt {4-12}}{2}$$\displaystyle =\frac{-2\pm \sqrt {-4\cdot 2}}{2}=-1 \pm i\sqrt{2}$
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  4. #4
    Senior Member Stroodle's Avatar
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    Awesome. THanks for your help guys
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  5. #5
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    Quote Originally Posted by Gamma View Post
    $\displaystyle \frac{z-1}{z+1}=z+2\Rightarrow z-1=z^2+3z+2 $

    How did you get from there-to-there, to-there?
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  6. #6
    MHF Contributor
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    Quote Originally Posted by Mr Smith View Post
    How did you get from there-to-there, to-there?
    Multiplying both sides of the equation by $\displaystyle z + 1$.
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