Solution Let: z1=a+bi,z2=c+di(a,b,c,d∈R)z_1=a+bi,\quad z_2=c+di\qquad (a,b,c,d\in\mathbb{R})z1=a+bi,z2=c+di(a,b,c,d∈R) Then: z1z2=(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+i(ad+bc)\begin{aligned} z_1z_2&=(a+bi)(c+di) \\ &=ac+adi+bci+bdi^2 \\ &=(ac-bd)+i(ad+bc) \end{aligned}z1z2=(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+i(ad+bc) So: z1z2‾=(ac−bd)−i(ad+bc)\overline{z_1z_2}=(ac-bd)-i(ad+bc)z1z2=(ac−bd)−i(ad+bc) Also: z1ˉ=a−bi,z2ˉ=c−di\bar{z_1}=a-bi,\quad \bar{z_2}=c-diz1ˉ=a−bi,z2ˉ=c−di z1ˉ z2ˉ=(a−bi)(c−di)=ac−adi−bci+bdi2=(ac−bd)−i(ad+bc)\begin{aligned} \bar{z_1}\,\bar{z_2}&=(a-bi)(c-di) \\ &=ac-adi-bci+bdi^2 \\ &=(ac-bd)-i(ad+bc) \end{aligned}z1ˉz2ˉ=(a−bi)(c−di)=ac−adi−bci+bdi2=(ac−bd)−i(ad+bc) Hence: z1z2‾=z1ˉ z2ˉ\boxed{\overline{z_1z_2}=\bar{z_1}\,\bar{z_2}}z1z2=z1ˉz2ˉ