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

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?

You don't just set $t=\frac 1z$ to compute a residue at infinity. The formula is $$\operatorname{Res}(f,\infty)=\operatorname{Res}\left(-\frac 1{z^2}f\left(\frac1z\right),0\right).$$
• You're welcome. It is interesting to figure out an intuitive explanation for this formula, in terms of which functions of the form $z^n$ have an anti-derivative and what happens to the orientation of the Riemann sphere under the transformation $z\mapsto\frac1z$. – Arnaud Mortier Jun 2 '18 at 17:44
• And one more question. is that $-{1\over z^2}$ is trivial in formula? Or it is specific for this problem? In my book it is just written $res(f,\infty) = -C_{-1}$ – Spike Bughdaryan Jun 2 '18 at 17:58