Circuit Theory/Phasors/j omega disappears

What makes $e^{j\omega }$ of $g(t)=\mathrm{Re}(G_m{e}^{j\varphi }{e}^{j\omega t})$ disappear?

First express voltage and current separately:

$v(t)=\mathrm{Re}(V_m{e}^{j\varphi }{e}^{j\omega t})$

$i(t)=\mathrm{Re}(I_m{e}^{j\varphi }{e}^{j\omega t})$

Every voltage in the Loop equations will have the $e^{j\omega }$ term in it so it can be canceled out.

$V_{m1}{e}^{j{\varphi }_1}{e}^{j\omega t}+V_{m2}{e}^{j{\varphi }_2}{e}^{j\omega t}+V_{m3}{e}^{j{\varphi }_3}{e}^{j\omega t}=V_s{e}^{j{\varphi }_s}{e}^{j\omega t}$

$V_{m1}{e}^{j{\varphi }_1}\cancel{e^{j\omega t}}+V_{m2}{e}^{j{\varphi }_2}\cancel{e^{j\omega t}}+V_{m3}{e}^{j{\varphi }_3}\cancel{e^{j\omega t}}=V_s{e}^{j{\varphi }_s}\cancel{e^{j\omega t}}$

$V_{m1}{e}^{j{\varphi }_1}+V_{m2}{e}^{j{\varphi }_2}+V_{m3}{e}^{j{\varphi }_3}=V_s{e}^{j{\varphi }_s}$

${𝕍}_{m1}+{𝕍}_{m2}+{𝕍}_{m3}={𝕍}_s$

Every current in the Junction (node) equations will have the $e^{j\omega }$ term in it so it can be canceled out.

$I_{m1}{e}^{j{\varphi }_1}{e}^{j\omega t}+I_{m2}{e}^{j{\varphi }_2}{e}^{j\omega t}+I_{m3}{e}^{j{\varphi }_3}{e}^{j\omega t}=I_s{e}^{j{\varphi }_s}{e}^{j\omega t}$

$I_{m1}{e}^{j{\varphi }_1}\cancel{e^{j\omega t}}+I_{m2}{e}^{j{\varphi }_2}\cancel{e^{j\omega t}}+I_{m3}{e}^{j{\varphi }_3}\cancel{e^{j\omega t}}=I_s{e}^{j{\varphi }_s}\cancel{e^{j\omega t}}$

$I_{m1}{e}^{j{\varphi }_1}+I_{m2}{e}^{j{\varphi }_2}+I_{m3}{e}^{j{\varphi }_3}=I_s{e}^{j{\varphi }_s}$

${𝕀}_{m1}+{𝕀}_{m2}+{𝕀}_{m3}={𝕀}_s$