How do you express 1/(x^2+x+1) 1x2+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))1x2+x+1=√3i3(x+12+√32i)−√3i3(x+12−√32i)
Explanation:
This expression can only be split down further if we use Complex coefficients...
x^2+x+1x2+x+1
= (x+1/2)^2+3/4=(x+12)2+34
= (x+1/2)^2-(sqrt(3)/2i)^2=(x+12)2−(√32i)2
= (x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)=(x+12−√32i)(x+12+√32i)
So:
1/(x^2+x+1) = A/(x+1/2-sqrt(3)/2i)+B/(x+1/2+sqrt(3)/2i)1x2+x+1=Ax+12−√32i+Bx+12+√32i
=(A(x+1/2+sqrt(3)/2i)+B(x+1/2-sqrt(3)/2i))/(x^2+x+1)=A(x+12+√32i)+B(x+12−√32i)x2+x+1
=((A+B)x+(A(1/2+sqrt(3)/2i)+B(1/2-sqrt(3)/2i)))/(x^2+x+1)=(A+B)x+(A(12+√32i)+B(12−√32i))x2+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))
