F(f)(ω)*=F(f*)(-ω)

$ \mathcal F(f)(\omega)^*=\frac1{\sqrt{2\pi}}\int_\R f^*(t)e^{i\omega t}\mathrm dt

$ =\frac1{\sqrt{2\pi}}\int_\R f^*(t)e^{-i(-\omega)t}\mathrm dt

$ = \mathcal F(f^*)(-\omega)

$ \therefore\mathcal F(f)(\omega)^*=\mathcal F(f^*)(-\omega)

$ \because\mathcal F^{-1}(F)(t)^*=\mathcal F(F)(-t)^*

$ = \mathcal F(F^*)(t)

$ = \mathcal F^{-1}(F^*)(-t)