## richyw Group Title exponential form of a complex number? one year ago one year ago

1. richyw

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

2. CliffSedge

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

3. satellite73

are $$R, \omega,L, C$$ some real constants?

4. richyw

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

5. 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)}

6. richyw

oops the, $$wC-1/wL$$ in the denominator of the above three is squared

7. CliffSedge

I think that's an ok way to go.

8. richyw

I am so unbelievably lost man. this sucks.

9. CliffSedge

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

10. CliffSedge

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

11. CliffSedge

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