Theory question about "j" imaginary unit (AC circuit analysis)
If you put a minus sign in front of the number "5" it becomes "-5". 

Try and look at this differently. Try thinking that it rotates the number "5" (tied to the origin by a piece of string of length 5) through 180 degrees to become "-5"

OK so far? Negative signs are the same as rotating thru 180 degrees...

Why not extend this further to produce something you can "stick" in front of a positive number that rotates it thru 90 degrees - in EE this is usually called "j" and it acts to rotate a value (about the origin) thru 90 degrees counter-clock wise i.e. if you did it twice (j*j) you'd get 180 degrees ("-").

From this gem of knowledge you can therefore say j*j = -1 therefore, j = \$\sqrt{-1}\$

Just as a minus sign can rotate any positive value thru 180 degrees it can rotate any vector or phasor thru 180 degrees. The same applies to the j operator - it rotates any vector or phasor thru 90 degrees counter clockwise.

EDIT - forgot part of question: -

substituting j into the impedance of a capacitor. Remember the basic formula for a capacitor is Q=CV and therefore differentiating the variables we get: -

\$ I = \dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

This tells us that for a sinewave applied voltage across a capacitor, the current will also be a sinewave but differentiated into a cosine like this: -



If you tried to calculate the impedance (V/I) of a capacitor from the V-I relationship you'd get into trouble because when I passes thru zero, V is NOT zero so you get infinities. If on the other hand you apply a "j" to bring current in phase with voltage the math works out fine - current and voltage are aligned and impedance based on instantaneous values of V/I makes sense.

I'm aware that you are just starting out so I've tried to keep this both accurate and simple (maybe too simple for some?).

If you look at the inductor, the "j" can be applied to the voltage to align it with the current hence "j" is in the numerator for inductive reactance and j is in the denominator for capacitive reactance. There are subtleties lying around here that hopefully will make sense as you learn more - it's actually no coincidence that "j" appears to "follow" omega when it comes to impedances - my explanation doesn't cover that and neither does your question!