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# Lamhe Hindi Movie Utorrent Free Download VERIFIED Yeh Lamhe Judaai Ke movie torrent download Lamhe full movie hindi dubbed download 720p hd Yeh Lamhe Judaai Ke movie preview Yeh Lamhe Judaai Ke movie download mp4 . Yeh Lamhe Judaai Ke movie trailer theQ: Residue of $f(z)=(z^2+2z+2)(z-1)^3$ I wish to find the residue of $f(z)=z^2+2z+2$ at $z=0$. I write $f(z)=z^2+(z+2)(z-1)^2=(z+2)^2\frac{(z-1)^2}{z}$, which should be equal to something like $-3z^2+18z$. However, it is also equal to $3z^2+(z+1)(z-1)^2$ and then, after simplification, I arrive at $6z-3$, which isn’t what I’m looking for. What have I done wrong here? EDIT: I have to find the Laurent series of $f$ about $0$, so I only have one pole, so the residue of $f(z)$ is $0$. A: \begin{align} f(z) &= (z^2 + 2z + 2)(z-1)^3 \\ &= z^2 + 2z + 2(z-1)^3 \\ &= z^2 + 2z + 2z^3 – 6z^2 + 9z – 3 \\ &= z^2 + 6z – 3 \end{align} A: The problem is you have $\frac{(z-1)^2}{z} = (z-1)(1+\frac1z)$ As such, you now have $$f(z) = (z+2)^2(z-1)(1+\frac1z)$$ If you have the Laurent series for $(z+2)^2$, that’s the first thing to do. Next, we can evaluate the sum of the residues at $0$ and $z-1$: \text{Res}(f(z),0) + \text{Res}(