Imaginary part $\frac{R+\jmath \omega L}{1+\jmath \omega LR-\omega^2CL}$

Question

Someone can help me with stupid algebraic steps. I can't find the imaginary part of the following expression.

$$Z=\frac{R+\jmath \omega L}{1+\jmath \omega LR-\omega^2CL}$$ $$\jmath=\sqrt{-1}$$ R,L,$\omega$,C are real constants.

Thank you so much for your help.

Answer 1

$$Z=\frac{R+wLi}{1-w^2CL+wLRi}\cdot\frac{1-w^2CL-wLRi}{1-w^2CL-wLRi}=$$$${}$$

$$=\frac{\left(R-w^2RCL+w^2L^2R\right)+\left(wL-w^3CL^2-wLR^2\right)i}{\left|1-w^2CL+wLRi\right|^2}$$

So the imaginary part is

$$\frac{wL-w^3CL^2-wLR^2}{\left|1-w^2CL+wLRi\right|^2}$$

Answer 2

The best approach is to multiply both the numerator and the denominator by the complex conjugate of the denominator. This way the denominator is turned into a strictly real number: \begin{equation} \begin{split} Z&=\frac{R+\jmath\omega L}{1-\omega^2CL+\jmath\omega LR}\times\frac{1-\omega^2CL-\jmath\omega LR}{1-\omega^2CL-\jmath\omega LR}\\ &=\frac{R-\omega^2CLR+\omega^2L^2R+\jmath(\omega L-\omega^3CL^2-\omega LR^2)}{(1-\omega^2CL)^2+(\omega LR)^2}, \end{split} \end{equation} so that the imaginary part of the expression is equal to \begin{equation} \frac{\omega L-\omega^3CL^2-\omega LR^2}{(1-\omega^2CL)^2+(\omega LR)^2}. \end{equation}