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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?

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

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    $\begingroup$ Thanks, I got it $\endgroup$ – Spike Bughdaryan Jun 2 '18 at 17:43
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    $\begingroup$ 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 $z\mapsto\frac1z$. $\endgroup$ – Arnaud Mortier Jun 2 '18 at 17:44
  • $\begingroup$ 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 $res(f,\infty) = -C_{-1}$ $\endgroup$ – Spike Bughdaryan Jun 2 '18 at 17:58
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    $\begingroup$ @SpikeBughdaryan It is not specific to this problem, it is the general definition of the residue at infinity. $\endgroup$ – Arnaud Mortier Jun 2 '18 at 17:59
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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.

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  • $\begingroup$ Thanks, I will take a note $\endgroup$ – Spike Bughdaryan Jun 2 '18 at 17:43

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