Thevenin impedance and Vab
I'd probably just use nodal analysis, with \$V_a\$ (+) on the left side and \$V_b\$ (-) on the right side, \$V_s\$ as your source voltage, and where I'd choose to make the bottom node ground or \$0\:\textrm{V}\$:

$$\begin{align*}
\frac{V_a}{40\:\Omega}+\frac{V_a}{-2 \textrm{j}}&amp;= \frac{V_s}{40\:\Omega}+\frac{0\:\textrm{V}}{-2 \textrm{j}} \\ \therefore V_a &amp;= \frac{\frac{V_s}{40\:\Omega}}{\frac{1}{40\:\Omega}+\frac{1}{-2 \textrm{j}}} = \frac{V_s}{1+\frac{40\:\Omega}{-2 \textrm{j}}}=\frac{60\cdot\left(1-20\textrm{j}\right)}{401} \\ \\ \frac{V_b}{5 \textrm{j}}+\frac{V_b}{20\:\Omega}&amp;= \frac{V_s}{5 \textrm{j}}+\frac{0\:\textrm{V}}{20\:\Omega} \\ \therefore V_b &amp;= \frac{\frac{V_s}{5 \textrm{j}}}{\frac{1}{5 \textrm{j}}+\frac{1}{20\:\Omega}}=\frac{V_s}{1+\frac{5 \textrm{j}}{20\:\Omega}}=\frac{240\cdot\left(4-\textrm{j}\right)}{17}
\end{align*}$$

Then, as you say, just subtract. Since the positive side is on the left, the value is:

$$\begin{align*}V_{ab}&amp;=V_a-V_b=\frac{60\cdot\left(1-20\textrm{j}\right)}{401}-\frac{240\cdot\left(4-\textrm{j}\right)}{17}=\frac{4740}{6817}\cdot\left(-81+16\textrm{j}\right)\\ &amp;\approx -56.321 + 11.12513\textrm{j},~~~~~ 57.41~\angle~168.8^\circ \end{align*}$$