How do you express #1/(x^2+x+1) # in partial fractions?
1 Answer
#1/(x^2+x+1)=(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))#
Explanation:
This expression can only be split down further if we use Complex coefficients...
#x^2+x+1#
#= (x+1/2)^2+3/4#
#= (x+1/2)^2-(sqrt(3)/2i)^2#
#= (x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)#
So:
#1/(x^2+x+1) = A/(x+1/2-sqrt(3)/2i)+B/(x+1/2+sqrt(3)/2i)#
#=(A(x+1/2+sqrt(3)/2i)+B(x+1/2-sqrt(3)/2i))/(x^2+x+1)#
#=((A+B)x+(A(1/2+sqrt(3)/2i)+B(1/2-sqrt(3)/2i)))/(x^2+x+1)#
Equating coefficients, we find:
#{ (A+B=0), (A(1/2+sqrt(3)/2i)+B(1/2-sqrt(3)/2i)=1) :}#
From the first equation we find
Substitute this in the second equation to get:
#Asqrt(3)i = 1#
Hence:
#{ (A = 1/(sqrt(3)i) = -sqrt(3)/3i), (B=sqrt(3)/3i) :}#
So:
#1/(x^2+x+1) = B/(x+1/2+sqrt(3)/2i)+A/(x+1/2-sqrt(3)/2i)#
#=(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))#
