[tex] \dfrac{8}{2+2i}=\dfrac{8}{2(1+i)}=\dfrac{4}{1+i}=\dfrac{4}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{4(1+i)}{1^2-i^2}\\\\=\dfrac{4(1+i)}{1-(-1)}=\dfrac{4(1+i)}{1+1}=\dfrac{4(1+i)}{2}=2(1+i)=2+2i\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\i=\sqrt{-1}\to i^2=-1 [/tex]
