Circuit Theory/Phasors/j omega disappears

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.


 * $$I_{m1} e^{j\phi_1}e^{j\omega t} + I_{m2} e^{j\phi_2}e^{j\omega t} + I_{m3} e^{j\phi_3}e^{j\omega t} = I_s e^{j\phi_s}e^{j\omega t}$$
 * $$I_{m1} e^{j\phi_1}\cancel{e^{j\omega t}} + I_{m2} e^{j\phi_2}\cancel{e^{j\omega t}} + I_{m3} e^{j\phi_3}\cancel{e^{j\omega t}} = I_s e^{j\phi_s}\cancel{e^{j\omega t}}$$
 * $$I_{m1} e^{j\phi_1} + I_{m2} e^{j\phi_2} + I_{m3} e^{j\phi_3} = I_s e^{j\phi_s}$$
 * $$\mathbb{I}_{m1} + \mathbb{I}_{m2} + \mathbb{I}_{m3} = \mathbb{I}_s$$