f(z) = (z²+1) / (z(z-1)(z-3)) Using Partial Fraction

The denominator has three distinct linear factors: $z$, $(z-1)$, and $(z-3)$.

$$\frac{z^2+1}{z(z-1)(z-3)} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-3}$$

Multiply both sides by $z(z-1)(z-3)$:

$$z^2+1 = A(z-1)(z-3) + B(z)(z-3) + C(z)(z-1)$$

To find $A$, set $z=0$:

$$(0)^2+1 = A(0-1)(0-3) + B(0)(0-3) + C(0)(0-1)$$ $$1 = A(-1)(-3)$$ $$1 = 3A$$ $$A = \frac{1}{3}$$

To find $B$, set $z=1$:

$$(1)^2+1 = A(1-1)(1-3) + B(1)(1-3) + C(1)(1-1)$$ $$2 = A(0)(-2) + B(1)(-2) + C(1)(0)$$ $$2 = -2B$$ $$B = -1$$

To find $C$, set $z=3$:

$$(3)^2+1 = A(3-1)(3-3) + B(3)(3-3) + C(3)(3-1)$$ $$9+1 = A(2)(0) + B(3)(0) + C(3)(2)$$ $$10 = 6C$$ $$C = \frac{10}{6} = \frac{5}{3}$$

Substitute the values of $A$, $B$, and $C$ back into the decomposition form:

$$\frac{z^2+1}{z(z-1)(z-3)} = \frac{\frac{1}{3}}{z} + \frac{-1}{z-1} + \frac{\frac{5}{3}}{z-3}$$ $$\frac{z^2+1}{z(z-1)(z-3)} = \frac{1}{3z} - \frac{1}{z-1} + \frac{5}{3(z-3)}$$

The partial fraction decomposition is:

$$\frac{1}{3z} - \frac{1}{z-1} + \frac{5}{3(z-3)}$$