Evaluate $\int\limits_{|z|=1}\frac{\sin{1/z}}{(z-2)^2}dz.$ This is a question from an old exam. The two integrals are seemingly similar, but the first one seems quite tedious compared to the other one. It seems, according to the prof solution, that the hard part in the first integral is computing the residue at $z=0.$


  Evaluate the intergals
  
  a) $$\int\limits_{|z|=1}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$
  
  b) $$\int\limits_{|z|=3}\frac{\sin{(1/z)}}{(z-2)^2} \ dz$$


In the first, we have $2$ poles, first one is $z_1=0$ and the second one is a pole of order $2$, which is $z_2=2.$ However, only $z_1$ is inside our unitcircle so we only need to compute $\text{Res}_{z_1}(f(z))$ and apply the residue theorem.

Since our function is of the form $f(z)=\frac{g(z)}{(z-z_1)^k}$ the residue is given by

$$\text{Res}_{z_1}(f(z))=\frac{g^{k-1}(z_1)}{(k-1)!},$$

and here $k=2$ and $(\sin(1/z))'=-\cos(1/z)/z^2$, so

$$\text{Res}_{z_1}(f(z))=-\frac{\cos\frac{1}{z_1}}{z_1^2}=...\text{this is the moment I realised I'm screwed.}$$

Howver, computing the residue at $z_2=2$ for the other integral I can use conventional methods, but not here.

Can someone break down the main difference between these integrals and show how to find the residue at $z_1=0?$

EDIT:

I need to solve $a)$ using Laurent series.
We substitute $w = 1/z$ and taking the explicit parametrization $z=e^{it}$ we see that the integration path changes the orientation, this brings a minus in front of the integral, so 
we have:
$$
\begin{aligned}
J_1 
&amp;= 
\int_{|z|=1}\frac{\sin(1/z)}{(z-2)^2} \; dz
\\
&amp;=
-\int_{|w|=1}\frac{\sin w}{\left(\frac 1w-2\right)^2} \; \left(-\frac 1{w^2}\right)\;dw
\\
&amp;=
\int_{|w|=1}\frac{\sin w}{(1-2w)^2} \; \;dw
\\
&amp;=2\pi\; i
\operatorname{Residue}_{w=1/2}
\frac{\sin w}{(1-2w)^2}
\\
&amp;=2\pi\; i
\cdot\frac 14\operatorname{Residue}_{w=1/2}
\frac
{\sin\left( \left(w-\frac 12\right)+\frac 12\right)}
{\left(w-\frac 12\right)^2}
\\
&amp;=2\pi\; i\cdot\frac 14\cdot\cos\frac 12
\ ,
\end{aligned}
$$
because in the expansion $\sin\left( \left(w-\frac 12\right)+\frac 12\right)
=
\sin\left(w-\frac 12\right)\cos\frac 12+
\cos\left(w-\frac 12\right)\sin\frac 12
$
only the $\sin\left(w-\frac 12\right)$ brings the needed odd power that contributes to the residue.

Let us check the above numerically, pari/gp:

? intnum( t=0, 2*Pi, sin( exp(-I*t) )  / ( exp(I*t)-2)^2 * I * exp(I*t) )

%1 = 2.2522481052807061131814259376142973690 E-58
   + 1.3785034646766524887891155026097526287*I
? Pi/2*cos(1/2.)
%2 = 1.3785034646766524887891155021962680546


(Result was manually rearranged.)

The other integral is converted by the same trick into one over $|w|=1/3$, and the contour avoids the pole $1/2$.
The integrand has a third-order zero at infinity, therefore the residue at infinity is $0$. Assuming that the contour goes around the origin counterclockwise, the first integral is $-2 \pi i$ times the residue at $z = 2$ and the second integral is $0$.