Help with AC circuit theory problem
Let's start with a few considerations on given schematics
  

First we are asked to relate \$ \Delta U_{12}\$ to switch closure, let's try to get them closer running along the highlighted branches:  

\$U_{12}=U_1-U_2=\$ but \$U_2=U_3+E_2-Z_1\,I_{g1}\$ and substituting  

\$U_{12}=U_1-U_3-E_2+Z_1\,I_{g1}\$ and going to variations  

\$\begin{align}\Delta U_{12} &amp;= U_{12}^{(c)}-U_{12}^{(o)} &amp;=U_1^{(c)}-U_3^{(c)}-E_2+Z_1\,I_{g1} &amp; \\
 &amp; &amp;- U_1^{(o)}+U_3^{(o)}+E_2-Z_1\,I_{g1} &amp;= \Delta U_{13}
\end{align}\$  

given constant \$E_2\$, \$I_{g1}\$ and obviously \$Z_1\$ we got to  


\$\Delta U_{12}= \Delta U_{13}\$
Now we remove \$Z_1\$ since any impedance series connected to a
current generator does not affect the network itself but only drop
across generator.
Then dualy we remove \$Z_2\$ for it is parallel connected to a
voltage generator and does not affect the network itself but only
current supplied by generator.
Now it is time to split current generator \$I_{g1}\$ into two
identical ones series connected, common point can be connected
anywhere we like since no current is going to flow thru.





Now left \$I_{g1}\$  founds itself parallel connected to a voltage
source and can hence be removed.
Finally we add \$E_2\$ and \$E_6\$ to get an equivalent \$E=E_6-E_2\$



On this circuit, given constant Ig1 it is easy to relate \$\Delta U_{13}\$ and \$\Delta U_{35}\$  variations using a simple voltage divider


$$\Delta U_{13}=-\frac{Z_5}{Z_5+Z_4}\Delta U_{35}$$


Now it still looks like we have two many unknowns (E and Ig1 compared to the single relation on \$\Delta U_{12}=\Delta U_{13}\$ we have) but this is not true. Our circuit may be thought driven by a single equivalent generator. 



where \$Z_e=Z_3 || (Z_4+Z_5)\$ and \$J\$ is a linear combination of E and Ig1 which we don't even need to calculate.

Now it's just plain sailing

\$\left\{\begin{align}
U_{35}^{(o)} &amp;= Z_e &amp; (J+I_{g2}) \\
U_{35}^{(c)} &amp;= (Z_e || Z) &amp; (J+I_{g2})
\end{align}\right. \$


$$\Delta U_{35}=(Z_e||Z-Z_e)(J+I_{g2})=-\frac{Z_e^2}{Z_e+Z}(J+I_{g2})$$


and given what found above in (7)


$$\Delta U_{12}=\Delta U_{13}=-\frac{Z_5}{Z_5+Z_4}\Delta U_{35}$$


and combining (8) and (9) we get total current driven to node 3


$$J+I_{g2}=\frac{(Z_5+Z_4)(Z_e+Z)}{Z_5Z_e^2}\Delta U_{12}$$


and hence \$U_{35}^{(c)}\$ which is needed to compute request Ig2 power.


$$U_{35}^{(c)}=(Z_e || Z)(J+I_{g2})=\frac{Z(Z_5+Z_4)}{Z_5Z_e}\Delta U_{12}$$
$$S_{Ig2}^{(c)}=U_{35}^{(c)}I_{g2}^*$$


Numerically we get




  Note: an error had crept into my line 8.
  
  and propagated till the end. Now it is fixed.