How do you write the following quotient in standard form #1/(4-5i)^2#?

1 Answer
Nov 13, 2017

Answer:

#-9/1681+40/1681i#

Explanation:

#"expand the denominator using FOIL method"#

#•color(white)(x)i^2=(sqrt(-1))^2=-1#

#rArr(4-5i)^2=16-40i+25i^2=-9-40i#

#rArr1/(4-5i)^2=1/(-9-40i)#

#"to rationalise the denominator multiply "#
#"numerator/denominator"#

#"by the "color(blue)"conjugate " "of "-9-40i#

#"the conjugate of "-9-40i" is "-9color(red)(+)40i#

#rArr(-9+40i)/((-9-40i)(-9+40i))#

#=(-9+40i)/1681#

#=-9/1681+40/1681ilarrcolor(blue)"in standard form"#