How do you write #(a-bi)/(c+di)# in standard form?
1 Answer
Dec 13, 2016
Explanation:
#(a-bi)/(c+di) = ((a-bi)(c-di))/((c+di)(c-di))#
#color(white)((a-bi)/(c+di)) = ((ac-bd)-(ad+bc)i)/(c^2+d^2)#
#color(white)((a-bi)/(c+di)) = (ac-bd)/(c^2+d^2)-(ad+bc)/(c^2+d^2)i#
