OpenCircuit Condition:  

The current source pushes 3 A of current through the lower 2 ohm resistor, thus there is a 6 V difference across it with polarity as shown, In the top portion, the 8 V source across the 2+j2 ohm impedance would create a current of 2j2 A through the top 2 ohm resistor; thus it has a voltage of 4j4 V across it. The sum of the two, 10j4 V is the Thevenin voltage. 
Killing the Sources:  

Shorting the voltage source and opening the current source yields this combination of passive elements. The j2 ohm and 2 ohm elements at the top are in parallel and have the equivalent impedance of 1+j ohm. The Thevenin impedance is thus 3+j ohm. 
Using the Generator:  

The total impedance of the equivalent Thevenin circuit is 3j ohm,
so the current through the resistor is
I = (10j4)/(3j) = (10.8\/21.8) / (3.16\/18.4) = 3.41\/3.4a 