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

3 Answers
Mar 31, 2017

Answer:

#(2i)/(5 -12i)#

Explanation:

#(1 + i)^2/(-3 + 2i)^2 = (1 + 2i + i^2)/(9 - 12i +4i^2)#

#i^2 = -1#, therefore

#(1 + 2i + i^2)/(9 - 12i +4i^2) = (1 + 2i +(-1))/(9 - 12i +4(-1))#

#= (1 + 2i -1)/(9 - 12i - 4) = (2i)/(5 -12i)#

Mar 31, 2017

Answer:

The answer is #color(red)((2i)/(5-12i))#.

Explanation:

According to problem(ATP),
#((1+i)^2)/((-3+2i)^2)#
=#(1-1+2i)/(4(i)^2-12i+9)#[as #i^2=-1#]
=#(2i)/(5-12i)#

Mar 31, 2017

Answer:

#(1+i)^2/(-3+2i)^2 = -24/169+10/169i#

Explanation:

Given:

#(1+i)^2/(-3+2i)^2#

Let us first simplify:

#(1+i)/(-3+2i)#

and then square it...

#(1+i)/(-3+2i) = ((1+i)(-3-2i))/((-3+2i)(-3-2i))#

#color(white)((1+i)/(-3+2i)) = (-3-2i-3i-2i^2)/(9-4i^2)#

#color(white)((1+i)/(-3+2i)) = (-3-2i-3i+2)/(9+4)#

#color(white)((1+i)/(-3+2i)) = (-1-5i)/13#

So:

#(1+i)^2/(-3+2i)^2 = ((1+i)/(-3+2i))^2#

#color(white)((1+i)^2/(-3+2i)^2) = ((-1-5i)/13)^2#

#color(white)((1+i)^2/(-3+2i)^2) = (-24+10i)/169#

#color(white)((1+i)^2/(-3+2i)^2) = -24/169+10/169i#