I am tasked to find the Laurent series of the function $f(z) = \frac{1 - 2z}{(z^2 - z)^2}$ on the annulus $\Delta^*(0,1) = \{z \in \mathbb{C}\mid 0 < |z| < 1\}$. The hint I have been given is to consider first the Laurent series of the function $g(z) = \frac{1}{z^2 - z}$. I have identified that $f'(z) = g(z)$ therefore suggesting that the termwise derivative of $f$'s Laurent series is $g$'s Laurent series on $\Delta^*(0,1)$. What I am unsure about is the convergence of $g$'s series on the said annulus. To be specific,
$$f(z) = \frac{1}{z}\frac{1}{z - 1} = \frac{-1}{z}\frac{1 - z} = \frac{-1}{z}\sum_{n=0}^\infty z^n = -\left(\frac{1}{z} + \sum_{n=0}^\infty z^n\right)$$$
Therefore,
$$g(z) = f'(z) = \frac{1}{z^2} - \sum_{n=0}^\infty nz^{n-1} = \frac{1}{z^2} - \sum_{n=1}^\infty nz^{n-1} = \frac{1}{z^2} - \sum_{n=0}^\infty (n + 1)z^n$$
Now, $\left|\frac{(n+2)z^{n+1}}{(n+1)z^n}\right| = \left|\frac{(n+2)z}{(n+1)}\right| < 1 \Longleftrightarrow 1 < \frac{n+2}{n+1} < |z|$ which surely does not hold on $\Delta^*(0,1)$.
How should I modify my approach or have I missed something?
As a bonus question: Is it true that the termwise derivative/integral of a Laurent series on a given annulus remains as a valid Laurent series after the said transformation?
Take my answer with a grain of salt because I did the calculations a bit fast. So, $$f(z)=\frac{1-2z}{z^2(1-z)^2}=\frac{1-z-z}{z^2(1-z)^2}=\frac{1-z}{z^2(1-z)^2}-\frac{z}{z^2(1-z)^2}=\frac{1}{z^2}\frac{1}{1-z}-\frac{1}{(1-z)^2}\frac{1}{z}.$$ So we use the geometric series here for $\frac{1}{1-z}$ and $\frac{1}{(1-z)^2}$ centered at 0 with $0<|z|<1.$ We get: $$\frac{1}{z^2}\sum_{n=0}^{\infty}z^n-\frac{1}{z}\sum_{n=1}^{\infty}nz^{n-1}$$ which is equal to: $$\frac{1}{z^2}\sum_{n=0}^{\infty}z^n-\frac{1}{z^2}\sum_{n=1}^{\infty}nz^{n}=\frac{1}{z^2}\left(\sum_{n=0}^{\infty}z^n-\sum_{n=1}^{\infty}nz^{n}\right)=\frac{1}{z^2}\left(\sum_{n=0}^{\infty}z^n-\sum_{n=0}^{\infty}nz^n\right)=\frac{1}{z^2}\sum_{n=0}^{\infty}(1-n)z^n=\sum_{n=0}^{\infty}(1-n)z^{n-2}.$$
Please check the calculations again, and let me know if I made any mistake. Hope it is alright mate, have a good one!