Circuit Theory/Phasors/proof2

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

$g(t)=G_{m}cos(\omega t-\frac{\pi }{2})$

$g(t)=G_m\mathrm{Re}(e^{j(\omega t-\frac{\pi }{2})})$

$g(t)=G_m\mathrm{Re}(e^{-j\ast \frac{\pi }{2}}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(G_m{e}^{-j\ast \frac{\pi }{2}}{e}^{j\omega t})$

$g(t)=\mathrm{Re}(𝔾e^{j\omega t})$

$𝔾=G_m{e}^{-j\ast \frac{\pi }{2}}=G_m(cos(-\frac{\pi }{2})+j\ast sin(-\frac{\pi }{2}))=-jGm$