Residues theorem, calculate the integrates

Question

Find the residue of $f(z)=\dfrac{(z-1)^3}{z(z+2)^3}$ at $z=\infty$.

In my refference they say that $\mathrm{Res}(f;\infty)=-\mathrm{Res}\left(\dfrac{1}{z^2}F(z);0\right)$, where $F(z)=f\left(\dfrac{1}{z}\right)$.

$$-\lim_{z\to 0}\frac{1}{z^2}F(z)=-\lim_{z\to 0}\frac{1}{z^2}\frac{(\frac{1}{z}-1)^3}{\frac{1}{z}(\frac{1}{z}+2)^3}=-\lim_{z\to 0}\frac{1}{z}\frac{(\frac{1}{z}-1)^3}{(\frac{1}{z}+2)^3}$$

But this seems to me to be in trouble, anyone can help?

Answer 1

Hint: $\text{Res}(f,-2)+\text{Res}(f,0)+\text{Res}(f,\infty)=0$

So $\text{Res}(f,\infty)=-\text{Res}(f,-2)-\text{Res}(f,0)$

The pole at $z=0$ is simple, and the pole at $z=-2$ is of order $3$.

Answer 2

If you insist on using the definition for residue at infinity, then $$f(z)=\frac{(z-1)^3}{z(z+2)^3}\implies -\frac{f(z)}{z^2}=-\frac{(z-1)^3}{z^3(z+2)^3}$$ Because it is a pole of order 3, you need to find $$\lim_{z\to0}\frac{1}{(3-1)!}\frac{d^{3-1}}{dz^{3-1}} \left[z^3\cdot\frac{-(z-1)^3}{z^3(z+2)^3}\right]=\lim_{z\to0}-\frac 12 \frac{d^2}{dz^2}\left[\frac{(z-1)^3}{(z+2)^3}\right]$$ Obtained by using the general formula for a residue.