$\textbf{Theorem:}$ Let $f$ be analytic in $|z–z_0|<R$ with $f(z_0)=w_0$, $f′(z_0)\neq 0 $ and $f(z)–w_0 \neq 0$ for $0<|z–z_0|\leq r \lt R$.Then $$g(w):=\frac{1}{2\pi i} \int_{C} \frac{z f'(z)}{f(z)-w}dz$$ where C is the circle $|z – z_0| = r$, is a holomorphic function for $|w – w_0| < m:=min_{C}|f – w_0|$. Moreover, it is the unique solution of $f(z) =w$ in $|w – w_0| < m$, so it is the local inverse of $f$ with the expansion $$g(w):=f^{-1}(w) = \sum_{n=0}^{\infty} \frac{1}{2\pi i} \int_{C} \frac{z f'(z)}{(f(z)-w_0)^{n+1}} (w-w_0)^{n}dz$$
Using the above theorem: Find the Maclaurin series for $$g(w):= \frac{1}{\pi i} \int_{C} \frac{z(z+1)}{z^2+2z-w}dz$$ and sum it in closed form.
My Approach
Clearly, $f(z)=z^2+2z$, and I can re-write the integral by multiplying and dividing by 2 so that it takes the form as mentioned in the theorem, which is $$g(w):= \frac{1}{2\pi i} \int_{C} \frac{2z(z+1)}{z^2+2z-w}dz$$
$\textbf{Question}:$ The choice for $C:|z-z_0|=r$. I need to choose a $z_0$ such that the above theorem holds. I took $z_0 =0$, and my $r=1$. By the above theorem, I get $$g(w):= \sum_{n=0}^{\infty}\big[ \frac{1}{2\pi i} \int_{C} \frac{2z(z+1)}{(z^2+2z)^{n+1}}dz \big] (w)^{n}$$
I can further use the Cauchy integral formula to find the coefficient of the series, that is $$\frac{f^{n-1}(0)}{(n-1)!}$$ where $f(z)=\frac{2(z+1)}{(z+2)^{n+1}}$
My question is whether I have the right approach or not. Secondly, what does it mean by saying write it in closed form and how to further proceed writing the simplified $g(w)$ in Maclaurin series.