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Math Help - Need help with a quadratic that has complex roots

  1. #1
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    Need help with a quadratic that has complex roots

    This is the equation in question:

    1-2r cos(\Theta) z^{-1}+r^2 z^{-2}=0


    I'm out of practice for these and don't remember how best to handle this. If I plug it into the quadratic equation, I get a square root that I'm not sure how to simplify. I'm guessing I need to apply Euler's Formula, but again, I'm out of practice.

    I know the answer to this is:

    z=re^{\pm j\theta}

    ...I'm just not sure how to get there.

    Can someone give me a hand?
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  2. #2
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    Re: Need help with a quadratic that has complex roots

    Hey Lancet.

    Try multiplying both sides by z^2 and consider anything not a z as a constant (if it is a constant). You then get something along the lines of az^2 + bz + c = 0.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Need help with a quadratic that has complex roots

    I would first multiply through by z^2 and write in standard form:

    z^2-2r\cos(\theta)z+r^2=0

    Apply the quadratic formula:

    z=\frac{2r\cos(\theta)\pm\sqrt{(-2r\cos(\theta))^2-4(1)(r^2)}}{2(1)}=\frac{2r(\cos(\theta)\pm\sqrt{ \cos^2(\theta)-1})}{2}=

    r\(\cos(\theta)\pm j\sin(\theta)\)

    Now, apply Euler's formula to obtain:

    z=re^{\pm j\theta}
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  4. #4
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    Re: Need help with a quadratic that has complex roots

    Quote Originally Posted by MarkFL2 View Post

    Apply the quadratic formula:

    z=\frac{2r\cos(\theta)\pm\sqrt{(-2r\cos(\theta))^2-4(1)(r^2)}}{2(1)}=\frac{2r(\cos(\theta)\pm\sqrt{ \cos^2(\theta)-1})}{2}=
    Ah! That's where I was getting stuck. I didn't know what to do with

    \sqrt{ \cos^2(\theta)-1}

    I didn't realize that was a modified trig identity. Thanks!
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