Calculation of a voltage in the frequency domain
The current source doesn't influence the circuit after the switch has closed because all the current it provides will flow through the switch itself, therefore it is not useful to the computation of \$V_c\$ in any way.

To see why, think of two impedances in parallel with the current source: one is the switch, the other is the equivalent of the circuit of all the other elements to the right of the switch. Then apply the current divider formula and you'll see that the zero-impedance (ideal) switch will hog all the current from the source.







simulate this circuit &ndash; Schematic created using CircuitLab

\[
I_x = I \cdot \dfrac {Z_{sw}} {Z_{sw} + Z_x}
\]

\$I_x = 0\$ if the switch is closed, i.e. if \$Z_{sw} = 0\$.