Question

Anybody knows why we put R = z-1 when finding poles? Thanks

Answer 1

dinkleburg

Answer 2

Every polynomial can be factored as a product of roots:
$$
z^n+z^{n-1}+\cdots z^2+z+c=(z-z_1)(z-z_2)\cdots(z-z_{n-1})(z-z_n)
$$
In this form, it is easy to see where the poles, \(p_i\\), and zeros, \(z_i\\), lay:
$$
f(z)=\frac{(z-z_1)(z-z_2)\cdots(z-z_{n-1})(z-z_n)}{(z-p_1)(z-p_2)\cdots(z-p_{n-1})(z-p_n)}
$$