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richyw Group Title

exponential form of a complex number?

  • 2 years ago
  • 2 years ago

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  1. richyw Group Title
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    how would I put \[Z=\frac{1}{\frac{1}{R}+i\left(\omega C - \frac{1}{\omega L}\right)}\] into its exponential form?

    • 2 years ago
  2. CliffSedge Group Title
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    I would first try to put it in the form z=a+bi if possible.

    • 2 years ago
  3. satellite73 Group Title
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    are \(R, \omega,L, C\) some real constants?

    • 2 years ago
  4. richyw Group Title
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    @CliffSedge I am trying to do that I multiplied by the complex conjugate and got \[Z=\frac{R-i\left(R^2(\omega C-\frac{1}{\omega L}\right)}{1+R^2\left( \omega C-\frac{1}{\omega L}\right)}\]

    • 2 years ago
  5. richyw Group Title
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    is this the right step? then \[a=\frac{R}{1+R^2\left( \omega C-\frac{1}{\omega L}\right)}\]and \[b=-\frac{R^2\left(\omega C-\frac{1}{\omega L}\right)}{1+R^2\left( \omega C-\frac{1}{\omega L}\right)}\] Z=\frac{R-i\left(R^2(\omega C-\frac{1}{\omega L}\right)}{1+R^2\left( \omega C-\frac{1}{\omega L}\right)}

    • 2 years ago
  6. richyw Group Title
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    oops the, \(wC-1/wL\) in the denominator of the above three is squared

    • 2 years ago
  7. CliffSedge Group Title
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    I think that's an ok way to go.

    • 2 years ago
  8. richyw Group Title
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    I am so unbelievably lost man. this sucks.

    • 2 years ago
  9. CliffSedge Group Title
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    What about what satellite asked, is everything besides the i a real number constant? If so, you can simplify the form with a substitution.

    • 2 years ago
  10. CliffSedge Group Title
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    If you let \[\large u=\frac{1}{R}\] and \[\large v=\omega C-\frac{1}{\omega L}\] then you can operate on \[\large z=\frac{1}{u+v i}\]

    • 2 years ago
  11. CliffSedge Group Title
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    That might make some of the algebra easier for the intermediate steps, then you can re-substitute at the end.

    • 2 years ago
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