I have to find current Ix. Equations: None My approach: 1) Turn 60<0 and 2+4j impedances (top left of circuit) into current source parallel to impedances. 2+4j = 4.47<63.43 Current source = 60<0 / 4.47<63.43 = 13.423 < 296.57 2) The two impedances 2 +4j are now in parallel with the 6 ohms, so I comebine those to get impedance Z1 = 3<36.9 3) Then I add the current source on the left to the current source on the right. So net current source is 13.423 < 296.57 + 5<90 = 9.23<310.56 4) Then I transform net current source and Z1 into voltage source in series with Z1, so voltage source is 27.69<347.46 I think that adding the current sources should not give me the wrong answer for Ix, but after solving it this way, I get Ix = 3.87<14.33. Apparantly this is wrong. The book did NOT add the current sources. After getting impedance Z1, it turned it back into voltage source in series with Z1. Then it calculated current Ix, and got 5.25<17.4. I know how to get the answer the book did, but I don't understand why my method (adding both current sources, then turning back into voltage source) doesn't give me the right answer!