Complex reactance formula explanation
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}\$