# Circuit Theory/Phasors/j omega disappears

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

First express voltage and current separately:

${\displaystyle v(t)=\operatorname {Re} (V_{m}e^{j\phi }e^{j\omega t})}$
${\displaystyle i(t)=\operatorname {Re} (I_{m}e^{j\phi }e^{j\omega t})}$

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

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

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

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