How do you simplify #(1 + 2i) / (3-4i) #?

1 Answer
Mar 24, 2018

Answer:

#-1/5+2/5i#

Explanation:

#"multiply the numerator/denominator by the "color(blue)"conjugate"#
#"of the denominator"#

#"this ensures that the denominator is real"#

#"the conjugate of "3-4i" is "3color(red)(+)4i#

#color(orange)"Reminder "color(white)(x)i^2=(sqrt(-1))^2=-1#

#rArr(1+2i)/(3-4i)xx(3+4i)/(3+4i)#

#=((1+2i)(3+4i))/((3-4i)(3+4i))larrcolor(blue)"expand using FOIL"#

#=(3+10i+8i^2)/(9-16i^2)#

#=(-5+10i)/25#

#=(-5)/25+10/25i#

#=-1/5+2/5ilarrcolor(red)"in standard form"#