Mesh Analysis
Solve the following two equations for I_{2}:
(2+j2)I_{1}  2I_{2} = 8
2 I_{1} + (4j2)I_{2} = 6
Life is short! I put this into MathCAD and found
I_{2} = 3.4+j0.2 A = 3.41 A \/ 3.4 deg
Nodal Analysis
Solve the following two equations for V_{1} and V_{2}:
Eq. 1: jV_{1} + V_{2} = 8
Eq. 2: V_{1} + (1+j)V_{2} = j6
Multiplying Eq. 1 times j and then adding to Eq. 2 yields:
(1+j2)V_{2} = j2,
thus
V_{2} = 0.8j0.4
(after conversion of the two complex numbers to polar form, division, then conversion back).
Back substitution into Eq. 1 yields:
V_{1} = 0.4j7.2
so that
V_{1}V_{2} = 0.4j6.8
Finally,
I_{2} = (V_{1}V_{2})/(j2) = (0.4j6.8)/(j2) = 3.4j0.2
Life is too short for that too!
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 