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richyw Group TitleBest ResponseYou've already chosen the best response.0
how would I put \[Z=\frac{1}{\frac{1}{R}+i\left(\omega C  \frac{1}{\omega L}\right)}\] into its exponential form?
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
I would first try to put it in the form z=a+bi if possible.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
are \(R, \omega,L, C\) some real constants?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
@CliffSedge I am trying to do that I multiplied by the complex conjugate and got \[Z=\frac{Ri\left(R^2(\omega C\frac{1}{\omega L}\right)}{1+R^2\left( \omega C\frac{1}{\omega L}\right)}\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
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{Ri\left(R^2(\omega C\frac{1}{\omega L}\right)}{1+R^2\left( \omega C\frac{1}{\omega L}\right)}
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
oops the, \(wC1/wL\) in the denominator of the above three is squared
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
I think that's an ok way to go.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I am so unbelievably lost man. this sucks.
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
What about what satellite asked, is everything besides the i a real number constant? If so, you can simplify the form with a substitution.
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
If you let \[\large u=\frac{1}{R}\] and \[\large v=\omega C\frac{1}{\omega L}\] then you can operate on \[\large z=\frac{1}{u+v i}\]
 one year ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.1
That might make some of the algebra easier for the intermediate steps, then you can resubstitute at the end.
 one year ago
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