Can someone explain the math behind the following relationships between the equalities in a capacitor's reactance?
$$X_C = \frac{1}{2\pi f C} = \frac{-j}{\omega C} = \frac{1}{j \omega C}$$
For instance the second member, $$\frac{1}{2\pi f C}$$ It has no complex component, how can be equal to the others when
$$2\pi f = \omega$$
The complex component is missing right?
And the two last ones, they are alike, but the complex component is tossed around a bit. I don't understand what is happening there either.
The individual expressions/members is fine, but according to my textbook they are supposed to be equal but, I don't see how.
$$$$
EDIT: The two last equalities are from my textbook, the second one is from the Electronic Tutorials webpage. All under the X_C symbol.
Link: http://www.electronics-tutorials.ws/filter/filter_1.html
Alright. So what I've derived from the comments, is that different definitions of $$X_{LC}\ \ and\ \ Z_{LC}$$ are used in different places. I was under the impression that X, in a mandatory way, always had the complex unit inside it and was always a pure imaginary quantity. Not true though. The examples in my textbook now makes more sense now because it would seem it uses Z as the imaginary part and X as the mixed im. and real part.
Best Answer
For a capacitor, there is the relation:
$$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}\tag1$$
Considering the voltage signal to be:
$$\text{V}_\text{C}\left(t\right)=\text{V}_\text{p}\sin\left(\omega t\right)\tag2$$
It follows that:
$$\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}=\omega\text{V}_\text{p}\cos\left(\omega t\right)\tag3$$
And thus:
$$\frac{\text{V}_\text{C}\left(t\right)}{\text{I}_\text{C}\left(t\right)}=\frac{\text{V}_\text{p}\sin\left(\omega t\right)}{\omega\text{C}\text{V}_\text{p}\cos\left(\omega t\right)}=\frac{\sin\left(\omega t\right)}{\omega\text{C}\sin\left(\omega t+\frac{\pi}{2}\right)}\tag4$$
This says that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is \$\frac{1}{\omega\text{C}}\$, and that the AC voltage lags the AC current across a capacitor by \$90\$ degrees (or the AC current leads the AC voltage across a capacitor by \$90\$ degrees).
This result is commonly expressed in polar form as:
$$\text{Z}_\text{c}=\frac{1}{\omega\text{C}}\cdot e^{-\frac{\pi}{2}\cdot\text{j}}\tag5$$
Or, by applying Euler's formula, as:
$$\text{Z}_\text{C}=-\text{j}\cdot\frac{1}{\omega\text{C}}=\frac{1}{\text{j}\omega\text{C}}\tag6$$
Now for \$\text{X}_\text{C}\$:
$$\text{X}_\text{C}=\left|-\text{j}\cdot\frac{1}{\omega\text{C}}\right|=\frac{1}{\omega\text{C}}\tag7$$
Where \$\omega=2\pi\text{f}\$