# 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.

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