How does $\frac{1}{2jw(1+jw)}$ become $\frac{-j(1-jw)}{2w(1+w^2)}$, where $j^2=-1$? 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.
 A: 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 :)
A: Here, it is quite apparent $j$ represents the imaginary unit ($j^2 = -1$). This is quite common usage in physics and electrical engineering. In mathematics, you should be using $i$ rather than $j$.
Anyway, the factor of $j$ in the denominator can be dealt with by noting that $\frac 1j = \frac j{j^2} = -j$, so you can "bring up" the $j$ and reverse the sign. Following this, you need to "realise" (in analogy to rationalise) the denominator by multiplying by the conjugate of $1+j\omega$, which is $1 - j\omega$. You can use the basic algebraic identity $(a+b)(a-b) = a^2 - b^2$ to see what this would work out to. Keep in mind that multiplying $j$ by itself would give you $-1$. If you carry out these simple steps, you should be able to verify the working.
A: 
si $j^{2}=-1$
$\dfrac{1}{2jw(1+jw)}=\dfrac{1}{2jw(1+jw)}\times\dfrac{j}{j}=\dfrac{1\times j}{2jw(1+jw)\times j}=\dfrac{j}{2j^{2}w(1+jw)}=\dfrac{j}{2(-1)w(1+jw)}=\dfrac{j}{-2w(1+jw)}=\dfrac{j}{-2w(1+jw)}\times \dfrac{1-jw}{1-jw)}=\dfrac{j\times (1-jw)}{-2w(1+jw)(1-jw)}=\dfrac{j\times (1-jw)}{-2w((1)^{2}-(jw)^{2})}=\dfrac{j\times (1-jw)}{-2w(1-(j)^{2}(w)^{2})}=\dfrac{j\times (1-jw)}{-2w(1-(-1)(w)^{2})}=\dfrac{j\times (1-jw)}{-2w(1+(w)^{2})}=\dfrac{-j\times (1-jw)}{2w(1+(w)^{2})}$

