#IMDIV("4+2i","3-i") =<math>\frac{4+2i}{3-i}*\frac{3+i}{3+i}</math> = <math>\frac{12+10i+2i^2}{3^2-i^2} = 10+\frac{10i}{10}</math> (because <math>i^2=-1</math>) = <math> 1+\frac{i}{1} = 1+i </math> | #IMDIV("4+2i","3-i") =<math>\frac{4+2i}{3-i}*\frac{3+i}{3+i}</math> = <math>\frac{12+10i+2i^2}{3^2-i^2} = 10+\frac{10i}{10}</math> (because <math>i^2=-1</math>) = <math> 1+\frac{i}{1} = 1+i </math> |