Question

(1+i)(1-i)
(3+2i)(3-2i)
($sqrt{3}$-i)($sqrt{3}$+I)
(5+I)(5-2i)
(4+2i)(4-i)

Answer 1

$(1+i)(1-i)=1-i^2=1+1=2 \ (3+2i)(3-2i)=9-4i^2=9+4=13 \ ( sqrt{3} -i)( sqrt{3} +i)=3-i^2=3+1=4 \ (5+i)(5-2i)-25-10i+5i-2i^2=25+2-5i=27-5i \ (4+2i)(4-i)=16-4i+8i-2i^2=16+2+4i=18+4i$