In this case, `X_L= 3\ Ω` and `X_C= 0` so `X_L- X_C= 3\ Ω`.

So in rectangular form, the impedance is written:

`Z = 5 + 3j\ Ω`

Using calculator, the **magnitude** of *Z* is given
by: `5.83`, and the angle `θ` (the phase difference) is given by:
`30.96^@`.

So the voltage leads the current by `30.96^@`, as shown in the diagram.

Presenting *Z* as a complex number (in polar form), we
have:

`Z = 5.83 ∠ 30.96^@\ Ω`.