Maclaurin/Laurent series of a complex log

Question

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?

Answer 1

Note that: \begin{equation} \frac{1}{z^2\text{Log}(1+2z)} = \frac 1{z^2} \cdot \frac 1{2z-2z^2+\frac{8z^3}{3}\ldots} = \frac 1{2z^3} \cdot \frac 1{1-(z-\frac{4z^2}{3}\ldots)} \end{equation} Expand using $\frac 1{1-z}=1+z+z^2+\ldots$.

The result will have negative powers, which is normal and part of a Laurent series.

Answer 2

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*}