Find the curve such that integral yields following values

Question

Here is an exam problem I've tried to solve. I've managed to solve the first part, but wasn't able to solve the second one. I've looked for it in textbooks, workbooks, but I couldn't find any guiding idea. In the second part I've tried changing the radius of the circle, but it all yields to same integral as in first part. Here is the problem: Let $$f(z) = \frac{5z^2 - 8}{z^3 - 2z^2}$$
(a) Calculate $\int_\Gamma f(z)dz$, where $\Gamma$ is a positively oriented unit circle.
(b) Find closed curves such that integrals of the function along those curves yield to $14\pi i$, $18\pi i$ and $-2\pi i$.
After decomposition of the rational function to sum of partial fractions, I've managed to solve $(a)$ smoothly; however, $(b)$ gives me headache.
Any hints or solutions are appreciated.

Answer 1

$$f(z)=\frac{5z^2-8}{z^2(z-2)}\;,\;\;f'(z)=\frac{-5z^3+24z-32}{z^2(z-2)^2}$$

so $\;z=0\;$ is a pole of order two and $\;z=2\;$ a pole of order one, and

$$Res_{z=0}(f)=\lim_{z\to0}z^2f'(z)=\lim_{z\to0}\frac{-5z^3+24z-32}{(z-2)^2}=\frac{-32}{4}=-8$$

$$res_{z=2}(f)=\lim_{z\to2}(z-2)f(z)=\lim_{z\to2}\frac{5z^2-8}{z^2}=\frac{12}4=3$$

and now apply Cauchy's Residue Theorem. For example, if $\;C\;'$ is a simple closed curve not passing through $\;z=0,2\;$ and containing in its interior only the pole $\;z=0\;$ , we get that

$$\oint_C f(z)\,dz=2\pi i(-8)=-16\pi i$$

If instead the above curve you pick a very similar curve winding around zero 8 times (or whatever) in the positive or negative sense, then...