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Math Help - Finding Real and Imaginary

  1. #1
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    Finding Real and Imaginary

    m*z^2 - D*z +k = 0<br />

    Need to find the values of z in terms of real and imaginary. m, D and k are all constants. Help would be much aprecciated!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by iphysics View Post
    m*z^2 - D*z +k = 0<br />

    Need to find the values of z in terms of real and imaginary. m, D and k are all constants. Help would be much aprecciated!
    Just find the discriminant !

    \Delta=D^2-4km

    The solutions are thus z=\frac{D \pm \sqrt{D^2-4km}}{2m}

    If D^2-4km <0, we can write D^2-4km=(-1)*\underbrace{(4km-D^2)}_{>0}=i^2 \cdot (4km-D^2)

    \implies z=\frac{D \pm i ~\sqrt{4km-D^2}}{2m}=\frac{D}{2m} \pm i ~\frac{\sqrt{4km-D^2}}{2m}

    What are the real and imaginary parts ?
    Last edited by Moo; August 27th 2008 at 09:55 AM. Reason: bad mistake :o
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  3. #3
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    Thanks for that man. Im going to scan the document in another topic, because it is near impossible for me to do. Will upload ASAP

    BTW, The real component is = to D/2m. Dont know how to explain the negative sign!?
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Just find the discriminant !

    \Delta=D^2-4km

    The solutions are thus z=\frac{{\color{red}-(-D)} \pm \sqrt{D^2-4km}}{2m}

    If D^2-4km <0, we can write D^2-4km=(-1)*\underbrace{(4km-D^2)}_{>0}=i^2 \cdot (4km-D^2)

    \implies z=\frac{{\color{red}-(-D)} \pm i ~\sqrt{4km-D^2}}{2m}=\frac{{\color{red}D}}{2m} \pm i ~\frac{\sqrt{4km-D^2}}{2m}

    What are the real and imaginary parts ?
    Quote Originally Posted by iphysics View Post
    Thanks for that man. Im going to scan the document in another topic, because it is near impossible for me to do. Will upload ASAP

    BTW, The real component is = to D/2m. Dont know how to explain the negative sign!?
    The explanation (in red) is simple ....

    By the way, I think you'll have better luck with your other post if you type it out ... People like myself rarely open unknown files, for obvious reasons.
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