Circuit Theory/Phasors/proof1

$g(t)=G_{m}cos(\omega t)$ starting point

$g(t)=G_m\mathrm{Re}(e^{j\omega t})$

$g(t)=G_m\mathrm{Re}(e^{j*0}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(G_m{e}^{j0}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(𝔾e^{j\omega t})$

$𝔾=G_m{e}^{j0}=G_m(cos(0)+j*sin(0))=Gm$