Contour integral $\oint_{|z|=1}\frac{z^2\sin(1/z)}{z-2}dz$

Solution 1:

Where are you getting $\pi i/6$? I am also getting a residue of

$$2-4\sin\left(\frac{1}{2}\right)$$

We can see this directly using power series centered at zero:

$$\frac{z^2\sin(1/z)}{z-2}=$$

$$z^2\cdot\left(\frac{1}{1!z}-\frac{1}{3!z^3}+\cdots\right) \cdot\left(\frac{-1/2}{1-z/2}\right)=$$

$$z^2\cdot\left(\frac{1}{1!z}-\frac{1}{3!z^3}+\cdots\right)\cdot \frac{-1}{2}\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\cdots\right)$$

We can now see term by term that the coefficient of $1/z$ will be:

$$\frac{-1}{2}\left(\frac{-1}{3!2^0z}+\frac{1}{5!2^2z}-\frac{1}{7!2^4z}+\cdots\right)$$

Which is clearly:

$$2-4\cdot\sin\left(\frac{1}{2}\right)$$