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Math Help - Square root of complex number

  1. #1
    Junior Member piglet's Avatar
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    Square root of complex number

    I'm at a loss how to go about solving equations like the one below, where there are j terms under the square root sign.



    I've tried getting the complex conjugate of what's under the square root sign but i'll still have a j term.

    I'm convinced it's a straightforward process and that I'm missing some simple trick.

    Clues/Hints/Ideas/Whatever welcome!
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  2. #2
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    Re: Square root of complex number

    Hello, piglet!

    I'm at a loss how to go about solving equations like the one below.

    \eta \,=\,\sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} \,=\,\sqrt{\frac{j\!\cdot\!2\pi(100 \times 10^6)(100)(4\pi \times 10^{-7})}{0.00964 + j\!\cdot\!2\pi(100 \times 10^6)(8.854 \times 10^{-12})}} \,=\,2664e^{j30}\Omega

    Did you do any of the arithmetic?


    The numerator is: . j\!\cdot\!2\pi (10^8)(10^2)(4\pi\times 10^{-7}) \;=\;8\pi^2j \times 10^3

    The denominator is: . (9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})
    . . . . . . . . . . . . . =\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4})
    . . . . . . . . . . . . . =\;(2\cdot10^{-4})(48.2 + 4.427\pi j


    The fraction becomes: . \frac{8\pi^2j \times 10^3} {(2\cdot 10^{-4})(48.2 + 4.427\pi j)}

    . . . . . . . . . . . . . . . =\;\frac{4\pi^2j \times 10^7}{48.2 + 4.427\pi j}



    By the way, what are we solving for?

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  3. #3
    Junior Member piglet's Avatar
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    Re: Square root of complex number

    Quote Originally Posted by Soroban View Post
    Hello, piglet!


    Did you do any of the arithmetic?


    The numerator is: . j\!\cdot\!2\pi (10^8)(10^2)(4\pi\times 10^{-7}) \;=\;8\pi^2j \times 10^3

    The denominator is: . (9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})
    . . . . . . . . . . . . . =\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4})
    . . . . . . . . . . . . . =\;(2\cdot10^{-4})(48.2 + 4.427\pi j


    The fraction becomes: . \frac{8\pi^2j \times 10^3} {(2\cdot 10^{-4})(48.2 + 4.427\pi j)}

    . . . . . . . . . . . . . . . =\;\frac{4\pi^2j \times 10^7}{48.2 + 4.427\pi j}



    By the way, what are we solving for?

    Yes I got it down that far but wondering how to solve that fraction when it's under a square root and then convert it to an polar exponential like the picture in my first post. Any ideas?

    By the way, we're solving for Eta, the intrinsic impedance of a medium that has a magnetic field passing through it.
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  4. #4
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    Re: Square root of complex number

    Quote Originally Posted by piglet View Post
    Yes I got it down that far but wondering how to solve that fraction when it's under a square root and then convert it to an polar exponential like the picture in my first post. Any ideas?

    By the way, we're solving for Eta, the intrinsic impedance of a medium that has a magnetic field passing through it.
    You multiply the top and bottom of the fraction by the complex conjugate of the denominator, which reduces the fraction to something of the form A+Bj

    Now convert that into polar form and the square root is easy-peasea from there.

     \sqrt{A+Bj}=\sqrt{\rho e^{i(\theta+2n\pi)}}=\sqrt{\rho}\ e^{i(\theta/2+n\pi)}};\ n=0, 1

    CB
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