 Solution

Factor the denominator as $(z-z_0)(z+z_0)$. Since $z_0\ne 0$, its zeros are distinct, and the potential simple poles and their residues areFor example, the first follows by taking ${\lim }_{z\to z_0}(z-z_0)f(z)$. If the numerator vanishes at either denominator zero, the singularity there is removable and the corresponding residue is zero. Thus the data identify the possible poles, not an unconditional pole at a point canceled by $g$.