Circuit Theory/Phasors/proof2

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

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

$g(t)=G_{m}\operatorname {Re} (e^{j(\omega t-{\frac {\pi }{2}})})$

$g(t)=G_{m}\operatorname {Re} (e^{-j*{\frac {\pi }{2}}}e^{j\omega t})$

$g(t)=\operatorname {Re} (G_{m}e^{-j*{\frac {\pi }{2}}}e^{j\omega t})$

$g(t)=\operatorname {Re} (\mathbb {G} e^{j\omega t})$

$\mathbb {G} =G_{m}e^{-j*{\frac {\pi }{2}}}=G_{m}(cos(-{\frac {\pi }{2}})+j*sin(-{\frac {\pi }{2}}))=-jGm$