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.

$I_{m1}e^{j\phi _{1}}e^{j\omega t}+I_{m2}e^{j\phi _{2}}e^{j\omega t}+I_{m3}e^{j\phi _{3}}e^{j\omega t}=I_{s}e^{j\phi _{s}}e^{j\omega t}$

$I_{m1}e^{j\phi _{1}}{\cancel {e^{j\omega t}}}+I_{m2}e^{j\phi _{2}}{\cancel {e^{j\omega t}}}+I_{m3}e^{j\phi _{3}}{\cancel {e^{j\omega t}}}=I_{s}e^{j\phi _{s}}{\cancel {e^{j\omega t}}}$

$I_{m1}e^{j\phi _{1}}+I_{m2}e^{j\phi _{2}}+I_{m3}e^{j\phi _{3}}=I_{s}e^{j\phi _{s}}$

$\mathbb {I} _{m1}+\mathbb {I} _{m2}+\mathbb {I} _{m3}=\mathbb {I} _{s}$