Question

exponential form of a complex number?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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)}

Answer 6

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

Answer 7

I think that's an ok way to go.

Answer 8

I am so unbelievably lost man. this sucks.

Answer 9

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

Answer 10

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

Answer 11

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