Consider the function $f(z)$ in the complex plane, which has a pole at $z = 0$, a cut from $z_*$ to $\infty$ and behaves like Log(z)$^{-1}$ for large $z$. In order to converge, one defines $g(z)$ such that
$$g(z) = \frac{f(z)}{z+z_0}$$
Now, function $g(z)$ has two poles, at $z = 0$ and $z = - z_0$. Applying Cauchy's integral formula,
$$g(z) = \frac{1}{2\pi i}\oint d\omega \frac{f(\omega)}{(\omega + z_0)(\omega-z)}$$
Blowing up the contour, I got that
$$g(z) =\frac{f(-z_0)}{z+z_0}+\frac{f(0)}{z_0 z}+ \frac{1}{2\pi i}\int_{z_*}^{\infty}\frac{2 i\, \text{Im}(f(\omega))}{(\omega + z_0)(\omega-z)}$$
Finally, $f(z)$ is obtained simply by multiplying by $(z+z_0)$. However, my poles seem kinda off from the proposed solution. What am I doing wrong?