How to find the partial fraction of $\frac{z^2-1}{(z+1) (z+2)}$?

Question

I tried the method of

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

but it didn't work. Why is this?

Answer 1

As both denominator and numerator have the same degree, first divide the polynomials:

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

and now do partial fractions on the second summand above.

Answer 2

your fraction is improper so first $$z^2 - 1 = (z-1)(z+1)$$ then you have $$\frac{z^2-1}{(z+1)(z+2)}=\frac{z-1}{z+2}$$ and then at last $$\frac{z-1}{z+2}=1-\frac{3}{z+2}$$