What will be Laurent series of $\frac{1}{z^2(z-1)} $ about z=0 and z= 1 ?

Around z = 0 , I have calculated as $\frac{1}{-z^2}{(1-z)^{-1}} $ = $\frac{1}{-z^2}{(1+z+z^2+...)} $

Around z = 1, I have calculated by putting $z-1 = t =>$ $\frac{1}{t}({1+t)}^{-2}$ = $\frac{1}{t}({1+t)}^{-2}$ and then expanding.

Is this the correct way to solve? Can I expand ${1+t}$ assuming $t < 1$ ?


Since $\frac1{z^2}=-\left(\frac1z\right)'$ and since\begin{align*}\frac1z&=\frac1{1-(1-z)}\\&=1+(1-z)+(1-z)^2+(1-z)^3+\cdots\\&=1-(z-1)+(z-1)^2-(z-1)^3+\cdots\end{align*}when $|z-1|<1$, you have$$\frac1{z^2}=1-2(z-1)+3(z-1)^2-4(z-1)^3+\cdots$$Therefore,$$\frac1{z^2(z-1)}=\frac1{z-1}-2+3(z-1)-4(z-1)^2+\cdots$$

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