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exponential form of… - QuestionCove
OpenStudy (richyw):

exponential form of a complex number?

4 years ago
OpenStudy (richyw):

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

4 years ago
OpenStudy (anonymous):

I would first try to put it in the form z=a+bi if possible.

4 years ago
OpenStudy (anonymous):

are \(R, \omega,L, C\) some real constants?

4 years ago
OpenStudy (richyw):

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

4 years ago
OpenStudy (richyw):

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)}

4 years ago
OpenStudy (richyw):

oops the, \(wC-1/wL\) in the denominator of the above three is squared

4 years ago
OpenStudy (anonymous):

I think that's an ok way to go.

4 years ago
OpenStudy (richyw):

I am so unbelievably lost man. this sucks.

4 years ago
OpenStudy (anonymous):

What about what satellite asked, is everything besides the i a real number constant? If so, you can simplify the form with a substitution.

4 years ago
OpenStudy (anonymous):

If you let \[\large u=\frac{1}{R}\] and \[\large v=\omega C-\frac{1}{\omega L}\] then you can operate on \[\large z=\frac{1}{u+v i}\]

4 years ago
OpenStudy (anonymous):

That might make some of the algebra easier for the intermediate steps, then you can re-substitute at the end.

4 years ago
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