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richyw
 3 years ago
exponential form of a complex number?
richyw
 3 years ago
exponential form of a complex number?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would first try to put it in the form z=a+bi if possible.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are \(R, \omega,L, C\) some real constants?

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

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0is 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)}

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0oops the, \(wC1/wL\) in the denominator of the above three is squared

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think that's an ok way to go.

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0I am so unbelievably lost man. this sucks.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about what satellite asked, is everything besides the i a real number constant? If so, you can simplify the form with a substitution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you let \[\large u=\frac{1}{R}\] and \[\large v=\omega C\frac{1}{\omega L}\] then you can operate on \[\large z=\frac{1}{u+v i}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That might make some of the algebra easier for the intermediate steps, then you can resubstitute at the end.
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