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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}(