# How does $\frac{1}{2jw(1+jw)}$ become $\frac{-j(1-jw)}{2w(1+w^2)}$, where $j^2=-1$

algebra-precalculuscomplex numbersfractions

I am trying not exactly to solve equation, but just change it from what is on right side to what is on left side. But I didn't do any math for years and can't remember what to.

$$\frac{1}{2jw(1+jw)}=\frac{-j(1-jw)}{2w(1+w^2)}$$

Here, $$j^2=-1$$.

If I am right, I should do something like this.
$$\frac{1}{(2jw(1+jw))}=\frac{(2jw(1-jw))}{(2jw(1+jw))}$$
But whatever I do I can't get what's shown above.

I will be thankful for any help.

So, correct me if I'm wrong, but this equation seems to come from some electrical physics context? The reason I'm saying that is because it seems that $$j$$ is the imaginary unit (such that $$j^2 = -1$$), which is usually called $$i$$, except for some areas of physics (to not confuse it with intensity)
If so, then a first step would be to nice that $$\dfrac{1}{j} = -j$$ (can you see why this is true?). And once you have that, you can do a common trick which is "multiplying by the conjugate".
Example: say you have an expression of the form $$\dfrac{1}{a + bj}$$, and you want to get a real denominator. You can achieve this by multiplying both the top and bottom by the conjugate of $$a + bj$$, which is defined as $$a - bj$$ (see complex conjugate, keeping in mind that the article uses $$i$$ instead of $$j$$). I will let you perform the calculations for yourself (because that will help you understand what's going on), but this should be enough for you to finish the computations :)