Find the real and imaginary part of $\sqrt{(R+i\omega L) (R C/L+i\omega C)}$

Question

I want to seperate the real and imaginary part of the following, but I'm stuck. $$ \gamma=\alpha+i\beta=\sqrt{(R+i\omega L)\Big (\frac{R C}{L}+i\omega C\Big)} $$ Attempt: \begin{align} \sqrt{(R+i\omega L)\Big (\frac{R C}{L}+i\omega C\Big)} &=\sqrt{\frac{R^2C}{L}+i\omega CR+i\omega RC+i^2 \omega^2LC}\\ &=\sqrt{\frac{R^2C}{L}+i\omega 2RC-\omega^2LC}\\ &=\sqrt{\frac{R^2C}{L}-\omega^2LC+i\omega 2 RC} \end{align} And I'm stuck here...

Answer 1

Hint: $$\frac {RC}L + i \omega C = \frac CL (R+i\omega L) $$

Answer 2

You can restart from you last step

$$ \alpha+i\beta =\sqrt{\frac{R^2C}{L}-\omega^2LC+i\omega 2 RC} =\sqrt{{C\over L}\left(R^2-\omega^2L^2+i\omega 2 RL\right)} $$ Square, then equal real parts and imaginary parts

$$\begin{align} &\alpha ^2 - \beta ^2 &=& \quad{C\over L}\left(R^2-\omega^2L^2\right) \tag 1\\ &2\alpha\beta &=& \quad 2\omega RC\tag 2\\ \end{align} $$ From $(2)$ we get $\beta=\omega RC/\alpha.$ Upload it into $(1)$ and solve the biquadratic equation $$\alpha ^4 - {C\over L}\left(R^2-\omega^2L^2\right)\alpha^2-\omega^2R^2C^2=0.$$ $\alpha, \beta$ are reals but I am not sure if they can be negative (in electronics or in other science), so I prefer to write both mathematically valid solutions: $$\alpha + i\beta =\pm \left(R\sqrt{C\over L}+i\omega\sqrt{CL}\right)$$