Answer by periblepsis for How to simplify this AC electrical circuit

You've put some work into thinking through some of the KCL and I can see better where you need help.

Here's the KCL:

$$\begin{align*}\frac{U_1}{5\:\Omega}+\frac{U_1}{j\,4}&=\frac{U_2}{j\,4}+2\:\text{A}\\\\\frac{U_2}{-j\,3}+\frac{U_2}{j\,4}&=\frac{U_1}{j\,4}+\left(2+j\,2\right)\:\text{A}\end{align*}$$

Note that I've included the capacitor and inductor currents. Do you see what I've done above and why?

This can be solved quickly into \$U_1\$ and \$U_2\$. And from those, you can get all the currents.

EDIT: Now that some time has passed (and I think you get the idea from the chat), I'll lay out what a Python/SymPy solver says about the above; for anyone else interested to test themselves:

solve([     Eq( U1/5 + U1/(4*I) , 2 + U2/(4*I) ),    Eq( U2/(-3*I) + U2/(4*I), U1/(4*I) + (2+2*I) )      ], [ U1, U2 ]){U1: 5 - 5*I, U2: 9 - 9*I}

So \$U_1=5\cdot\left(1-j\right)\$ and \$U_2=9\cdot\left(1-j\right)\$. The difference between these two lay across the inductor. So you can take the difference and divide by the inductive impedance to get its current. And so on for the other two parts.