Solve f(z) = z / ((z-1)(z+1)(z+2)) using Partial Fraction

The denominator has three distinct linear factors: $(z-1)$, $(z+1)$, and $(z+2)$. So, we can write the function as:

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

Multiply both sides by $(z-1)(z+1)(z+2)$:

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

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

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

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

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

To find $C$, set $z=-2$:

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

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

$$\frac{z}{(z-1)(z+1)(z+2)} = \frac{\frac{1}{6}}{z-1} + \frac{\frac{1}{2}}{z+1} + \frac{-\frac{2}{3}}{z+2}$$ $$\frac{z}{(z-1)(z+1)(z+2)} = \frac{1}{6(z-1)} + \frac{1}{2(z+1)} - \frac{2}{3(z+2)}$$

The partial fraction decomposition is:

$$\frac{1}{6(z-1)} + \frac{1}{2(z+1)} - \frac{2}{3(z+2)}$$