Circuit Theory/Phasors/j omega disappears

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What makes e^{j \omega} of g(t)=\operatorname{Re}(G_m e^{j\phi}e^{j\omega t}) disappear?

First express voltage and current separately:

v(t)=\operatorname{Re}(V_m e^{j\phi}e^{j\omega t})
i(t)=\operatorname{Re}(I_m e^{j\phi}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\phi_1}e^{j\omega t} + V_{m2} e^{j\phi_2}e^{j\omega t} + V_{m3} e^{j\phi_3}e^{j\omega t} = V_s e^{j\phi_s}e^{j\omega t}
V_{m1} e^{j\phi_1}\cancel{e^{j\omega t}} + V_{m2} e^{j\phi_2}\cancel{e^{j\omega t}} + V_{m3} e^{j\phi_3}\cancel{e^{j\omega t}} = V_s e^{j\phi_s}\cancel{e^{j\omega t}}
V_{m1} e^{j\phi_1} + V_{m2} e^{j\phi_2} + V_{m3} e^{j\phi_3} = V_s e^{j\phi_s}
\mathbb{V}_{m1} + \mathbb{V}_{m2} + \mathbb{V}_{m3} = \mathbb{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\phi_1}e^{j\omega t} + I_{m2} e^{j\phi_2}e^{j\omega t} + I_{m3} e^{j\phi_3}e^{j\omega t} = I_s e^{j\phi_s}e^{j\omega t}
I_{m1} e^{j\phi_1}\cancel{e^{j\omega t}} + I_{m2} e^{j\phi_2}\cancel{e^{j\omega t}} + I_{m3} e^{j\phi_3}\cancel{e^{j\omega t}} = I_s e^{j\phi_s}\cancel{e^{j\omega t}}
I_{m1} e^{j\phi_1} + I_{m2} e^{j\phi_2} + I_{m3} e^{j\phi_3} = I_s e^{j\phi_s}
\mathbb{I}_{m1} + \mathbb{I}_{m2} + \mathbb{I}_{m3} = \mathbb{I}_s