How do you express (x-1)/(1+x^2)x−11+x2 in partial fractions?
2 Answers
As the denominator
This is already in irreducible form using Real coefficients, but with Complex coefficients we have:
(x-1)/(1+x^2) = (1+i)/(2(x-i))+(1-i)/(2(x+i))x−11+x2=1+i2(x−i)+1−i2(x+i)
Explanation:
If we stick with Real coefficients, then this rational expression is already in irreducible form: The denominator has no linear factors with Real coefficients and the numerator is linear.
If we use Complex coefficients, then we can factor the denominator as:
1+x^2 = (x-i)(x+i)1+x2=(x−i)(x+i)
So we have:
(x-1)/(1+x^2) = A/(x-i)+B/(x+i)x−11+x2=Ax−i+Bx+i
color(white)((x-1)/(1+x^2)) = (A(x+i)+B(x-i))/(1+x^2)x−11+x2=A(x+i)+B(x−i)1+x2
color(white)((x-1)/(1+x^2)) = ((A+B)x+i(A-B))/(1+x^2)x−11+x2=(A+B)x+i(A−B)1+x2
Equating coefficients, we have:
{ (A+B = 1), (i(A-B) = -1) :}
Hence:
A-B = (-1)/i = i
Adding the first equation to this one, we have:
2A = 1+i
Hence:
A=1/2+1/2i
and:
B=1/2-1/2i
So:
(x-1)/(1+x^2) = (1+i)/(2(x-i))+(1-i)/(2(x+i))