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Math Help - Complex Equations

  1. #1
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    Complex Equations

    I'm gonna need a bit help here fellas.

    The equation is

    iz^2+(1+i)z-7+4i=0
    iz^2+(1+i)z-7+4i=0 - Wolfram|Alpha

    Can someone guide me through this one?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Complex Equations

    You see you have a quadratic (complex) equation in function of z, so just solve it like solving an other quadratic equation.
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  3. #3
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    Re: Complex Equations

    Hello, cristianodoni72!

    \text{Solve for }z\!:\;\;iz^2+(1+i)z-(7-4i)\:=\:0

    Quadratic Formula:
    . . z \;=\;\frac{\text{-}(1+i) \pm \sqrt{(1+i)^2 + 4(i)(7-4i)}}{2(i)} \;=\; \frac{\text{-}(1+i)\pm\sqrt{16 + 30i}}{2i}

    It happens that: . 16 + 30i \:=\:(5+3i)^2

    So we have: . z \;=\;\frac{\text{-}(1+i) \pm(5+3i)}{2i}

    The two answers are: . \begin{Bmatrix}\dfrac{\text{-}(1+i) + (5+3i)}{2i} &=&\dfrac{4+2i}{2i} &=& \dfrac{2}{i} + 1 &=& 1 - 2i \\ \\[-3mm] \dfrac{\text{-}(1+i) - (5+3i)}{2i} &=& \dfrac{\text{-}6-4i}{2i} &=& \text{-}\dfrac{3}{i} - 2 &=& \text{-}2 + 3i \end{Bmatrix}

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  4. #4
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    Re: Complex Equations

    Thanks alot! =) Now I get it!
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  5. #5
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    Re: Complex Equations

    Yes, now that Soroban did it for you, I'm sure you "get it". If he had not told you, would you have recognised that \sqrt{16+ 30i}= \pm (5+ 3i)?
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