PG,
You have to take into account the voltage across the current source, call it Vi. So the mesh equations are Vi-(2j*2j)+2*I1-24=0,2*I2+(-2j*I2)+(2j*2j)-Vi=0,I2-I1=2j. Solving for I2 gives 22/5+j*16/5. Multiplying by the 2 ohm output resistor gives 44/5+j*32/5 = 10.88/_36°.
Opening up the circuit at the left side of the capacitor gives a Voc of (24+(2*2j) and a Zo of 2 ohms. Using the voltage divider formula we get (24+(2*2j)*{2/(2+(-2j)+2)} = 10.88/_36°.
Ratch