# Verifying that $\sqrt{(R+j\omega L)(j\omega C)}=0.5\frac{R}{\sqrt{L/C}}+j\omega\sqrt{LC}$ for $R\lll \omega L$

So I have the following square root of this two complex numbers and my book provides this:

$$\sqrt{(R+j\omega L)(j\omega C)}=0.5\frac{R}{\sqrt{\frac{L}{C}}}+j\omega\sqrt{LC}$$

if $$R\lll\omega L$$

I have no freaking idea how they do this mathematically. I tried to apply distributive property, which leads to

$$\sqrt{jR\omega C-\omega^2LC}$$

And the second term of my expansion kind of looks like the second term of the expression $$\sqrt{-\omega^2LC}=j\omega\sqrt{LC}$$

But I don't if (a) this is correct and (b) how do I get the first term.

The basic idea is that $$\sqrt{1+x} \approx 1+\frac x2$$ when $$x \ll 1$$. The right side is the first two terms of the Taylor series. If you expand the left you have $$\sqrt{(R+j\omega L)(j\omega C)}=\sqrt{Rj\omega C-\omega^2LC}\\ =j\omega\sqrt{LC}\sqrt {R\frac 1{j\omega L}+1}\\ \approx j\omega \sqrt{LC}\left(1+\frac {R}{2j\omega L}\right)\\ =j\omega \sqrt{LC}+\frac R2\sqrt{\frac {C}{L}}$$ They owe you an approximation sign when they do the Taylor series step.
• @Matthias: Actually the $j$ indicates it is probably engineering. I was a physics student and remember getting the approximation signs, but it was many years ago. – Ross Millikan Dec 16 '18 at 18:23