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Find the residue of $\dfrac{z^2}{(z-1)(z-2)(z-3)}$ at $\infty$.

We know that $\text{Res} (f)_\infty +\text{Res} (f)_{\text{ at other poles}}=0$

Now $f$ has poles at $1,2,3$ of order $1$.

Sum of residues of $f$ at $1,2,3=\dfrac{1}{2}+(-4)+\dfrac{9}{2}=1\implies \text{Res} (f)_\infty =-1$ but the answer is not matching .

Where am I wrong?

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Your result is correct. Note that residue at infinity can be evaluated also in this way: $$\text{Res}(f,\infty)=-\text{Res}(f(1/z)/z^2;0)= \text{Res}\left(\frac{1}{z(z-1)(2z-1)(3z-1)};0\right)=-1.$$

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Then the person who wrote the answer made a mistake, because your answer is correct.

There is another way of computing the residue at $\infty$. It is equal to$$-\operatorname{res}_{z=0}\left(\frac{f\left(\frac1z\right)}{z^2}\right)=-\operatorname{res}_{z=0}\left(\frac{1}{-6 z^4+11 z^3-6 z^2+z}\right),$$but this is $-1$ once again.

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Put

$$f(z)=\dfrac{z^2}{(z-1)(z-2)(z-3)}\implies f\left(\frac1z\right)=\frac{\frac1{z^2}}{\left(\frac1z-1\right)\left(\frac1z-2\right)\left(\frac1z-3\right)}= $$

$$=\frac z{(1-z)(1-2z)(1-3z)}\implies\;\text{ the wanted residue is}:$$

$$-Res_{z=0}\left(\frac1{z^2}f\left(\frac1z\right)\right)=-\lim_{z\to0}\frac z{z(1-z)(1-2z)(1-3z)}=-1$$

and your answer is correct.

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