In the network in the figure with $V_{M1} = 8 V$, $\phi_{V1} = 100 deg$, $V_{M2} = 7 V$, $\phi_{V2} = 135 deg$, $I_M = 20 A$, $\phi_I = 95 deg$, $R_1 = 6 \Omega$, $R_2 = 15 \Omega$, and $\textbf{Z}_C = -7i \Omega$, find $\textbf{V}_o$in the circuit using source transformation.

There are 3 sources that you could transform.
Choosing $V_{M2}$ will result in losing the voltage across $R_2$.
Choosing $I_M$ and $R_1$ will result in a series voltage source-impedance combination in parallel with $V_{M1}$ and $Z_C$ which does not lead to a circuit simplification.
$V_{M1}$ will result in a current source and impedance in parallel to a voltage source in series with an impedance to find $V_o$. ** Be careful of signs.
Enter complex numbers as either a+bi, i.e., 3+4i, or
ae$\wedge$(bi) where b is in radians, i.e., 5e$\wedge$(0.927i).
The transformation of $V_{1}$ and $Z_C$ results in a current source (pointing up) of $\textbf{I}_1$ = A
Combining $\textbf{Z}_C$ and $R_1$ in parallel is $\textbf{Z}_3$ = $\Omega$
Combining $\textbf{I}_1$ and $I_M$ results in (pointing up) $\textbf{I}_3$ = A
The transformation of $\textbf{I}_3$ and $\textbf{Z}_3$ results in a voltage source of $\textbf{V}_3$ = V
The final value $\textbf{V}_o$ = V