Answer: To find the partial fraction decomposition of
f(x) = (x^3-4x^2+9x-5)/(x^2-2x+3)^2
We can start by factoring the denominator into irreducible quadratic factors:
x^2 - 2x + 3 = (x - 1 + 2i)(x - 1 - 2i)
where i is the imaginary unit.
Since the denominator has repeated irreducible quadratic factors, we can write the partial fraction decomposition as:
f(x) = A/(x - 1 + 2i) + B/(x - 1 - 2i) + C/(x - 1)^2
where A, B, and C are constants that we need to solve for.
To find A, we can multiply both sides by (x - 1 + 2i)^2 and then substitute x = 1 - 2i, which gives:
A = lim_(x -> 1 - 2i) [(x - 1 + 2i)^2 f(x)]
= (1 - 1 + 2i + 2i)^2 [(1 - 2i)^3 - 4(1 - 2i)^2 + 9(1 - 2i) - 5]/[(1 - 2i - 1 + 2i)^2(1 - 1 + 2i)^2]
= -2 - 3i
Similarly, to find B, we can multiply both sides by (x - 1 - 2i)^2 and then substitute x = 1 + 2i, which gives:
B = lim_(x -> 1 + 2i) [(x - 1 - 2i)^2 f(x)]
= (1 - 1 - 2i + 2i)^2 [(1 + 2i)^3 - 4(1 + 2i)^2 + 9(1 + 2i) - 5]/[(1 + 2i - 1 - 2i)^2(1 - 1 - 2i)^2]
= -2 + 3i
To find C, we can multiply both sides by (x - 1)^2 and then substitute x = 1, which gives:
C = lim_(x -> 1) [(x - 1)^2 f(x)]
= (1 - 1)^2 [(1)^3 - 4(1)^2 + 9(1) - 5]/[(1 - 1)^2(1 - 2i)^2(1 + 2i)^2]
= 1
Therefore, the partial fraction decomposition of f(x) is:
f(x) = (-2 - 3i)/(x - 1 + 2i) + (-2 + 3i)/(x - 1 - 2i) + 1/(x - 1)^2
Note that there are different ways to write the partial fraction decomposition, depending on how you choose to combine the complex constants.