Simplify #(2-i)/(1+i)^2# and write the answer in form of a + bi?

Question

1322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-26T12:46:47

#= - 1/2- i [a=-1/2 , b= -1]#

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

#= (2 -i) /(cancel 1+2 i -cancel 1) ; [i^2=-1] #

#= (2 -i) /(2 i) #

#= (i(2 -i)) /(2 i *i) #

#= (2 i -i^2) /(2 i^2) #

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

#= - 1/2- i [a=-1/2 , b= -1]# [Ans]