Finding the residue of function $\frac {\sin z}{z^2}$

Question

Problem is to find: $res[{\sin z \over z^2}, z = \infty]$

Answer is $-1$

Because I need at infinity, I changed variable $t = {1 \over z}$, and found it at $t=0$.

$$t^2\sin {1\over t} = t^2\bigg({1\over t}- {1\over t^3*3!} + {1\over t^5*5!} -... \bigg) = t - {1 \over t*3!} + {1 \over t^3*5!} - ...$$ and so residue is $-{1\over 3!}$

Where is my mistake?

Answer 1

You don't just set $t=\frac 1z$ to compute a residue at infinity. The formula is $$\operatorname{Res}(f,\infty)=\operatorname{Res}\left(-\frac 1{z^2}f\left(\frac1z\right),0\right).$$

Answer 2

Notice that the residue at infinity is the sum of the residues at finite value, changing sign.

The problem is done calculating the residue of the function in its only pole.