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Math Help - Complex Number (Modulus/Phase)

  1. #1
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    Complex Number (Modulus/Phase)

    I've been working with the equation below for too long that I'm forgetting the simplest operation. So...

    \frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2}

    Can the equation above by split into:

    \frac{z\theta_0^2}{\theta_0^2} - \frac{z\theta_0^2}{\theta^2}+ \frac{z\theta_0^2}{2i\theta\theta_0\phi}?
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  2. #2
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    Quote Originally Posted by Air View Post
    \frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2}
    Can the equation above by split into:
    \frac{z\theta_0^2}{\theta_0^2} - \frac{z\theta_0^2}{\theta^2}+ \frac{z\theta_0^2}{2i\theta\theta_0\phi}?
    NO!
    It is that same as: \frac{a}{b+c} \not= \frac{a}{b} + \frac{a}{c}
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  3. #3
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    Can I have hints on how to find the modulus?

    I've been told that \frac{z\theta_0^2}{-\theta^2+ 2i\theta\theta_0\phi+\theta_0^2} is a real number.
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    Can we write this as \frac{{zA^2 }}{{ - B + 2iBAC + A^2 }}
    where A,B,C are real numbers?
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    Quote Originally Posted by Plato View Post
    Can we write this as \frac{{zA^2 }}{{ - B + 2iBAC + A^2 }}
    where A,B,C are real numbers?
    Multiplying the numerator and the denominator by the conjugate of the denominator can get real and imaginary components. Then we can:

    \sqrt{(R)^2+(I)^2} where R is real number and I is the imaginary number co-efficient.

    Would this work?
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  6. #6
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    This is true.
    \frac{{zA^2 }}{{ - B + 2iBAC + A^2 }} = \frac{{zA^2 \left[ { - B - 2iBAC + A^2 } \right]}}{{\left( {A^2  - B} \right)^2  + 4A^2 B^2 C^2 }}
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  7. #7
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    Quote Originally Posted by Plato View Post
    This is true.
    \frac{{zA^2 }}{{ - B + 2iBAC + A^2 }} = \frac{{zA^2 \left[ { - B - 2iBAC + A^2 } \right]}}{{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 }}
    And so...

    \frac{zA^2 \left[ { - B + A^2 } \right]}{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 } - \frac{zA^2 \left[ 2BAC\right]}{\left( {A^2 - B} \right)^2 + 4A^2 B^2 C^2 } i
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