Maclaurin/Laurent series of a complex log I've been asked to find the Maclaurin series of
\begin{equation}
\text{Log}(1+2z)
\end{equation}
Which I've done, and found to be $2z-2z^2+\frac{8z^3}{3}...$,
I'm now told 'hence find the first three terms about $z=0$ of the function
\begin{equation}
\frac{1}{z^2\text{Log}(1+2z)}
\end{equation}
I see the second function is the reciprocal of the first multiplied by $\frac{1}{z^2}$ but I'm not sure how their series are related?

Hint:
\begin{align*}
\frac{1}{{z^2 \log (1 + 2z)}} & = \frac{1}{{2z^3 }}\frac{{2z}}{{\log (1 + 2z)}} = \frac{1}{{2z^3 }}\frac{{2z}}{{2z - 2z^2  + \frac{{8z^3 }}{3} -  \cdots }} = \frac{1}{{2z^3 }}\frac{1}{{1 - z + \frac{{4z^2 }}{3} -  \cdots }} \\ & = \frac{1}{{2z^3 }}\left( {1 - \left( { - z + \frac{{4z^2 }}{3} -  \cdots } \right) + \left( { - z + \frac{{4z^2 }}{3} -  \cdots } \right)^2  -  \cdots } \right) =  \cdots .
\end{align*}
