Thevenin equivalent circuit with current and voltage source
The Thevenin voltage looking into the terminals A &amp; B can be found by inspection using superposition.

\$V_{th} =  (-j2A) \cdot (-j4\Omega) + j4V \$

The first term is found by zeroing the voltage source on the right and the second term is found by zeroing the current source on the left.

The Thevenin impedance is easy too; simply zero both sources to get:

\$ Z_{th} = -j4 \Omega\$

The current is then found to be:

\$I_{resistor} = \dfrac{V_{th}}{Z_{th} + 8 \Omega}\$

I highly recommend reviewing the late Dr. Leach's notes on using superposition.  I've taught it this way for years.  If you practice using superposition, you will amaze your friends and professors alike by solving many circuits by inspection.