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I need to find the Taylor expansion at $z=0$ of $f(z)=\ln\frac{1-z^3}{1+z^3}$

I had two approaches:

The first,

$f(z)=\ln\frac{1-z^3}{1+z^3} = \ln(1-z^3) - \ln(1+z^3)$ but I realized that this is not true in the complex case.

The second,

$f(z)=\ln\frac{1-z^3}{1+z^3} = \ln(1+\frac{-2z^3}{1+z^3})$ and continue with the "normal" taylor expansion but this doesn't seen to get me to the solution.

I'd be happy for an idea.

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    I just checked in Mathematica, and the branch cuts of both forms coincide, so the identity is true even with the branch cut at ${z:z\leq0}$, but for |z|<1, you will not hit any branch cuts. Also, it may help to note that $\log\frac{1-z^3}{1+z^3}=-2\operatorname{arctanh}(z^3)$.2012-12-12

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If $z\in U(0),$ then $\dfrac{1-z^3}{1+z^3}\in U(1),$ thus it is possible to choose the regular branch of $\ln {\dfrac{1-z^3}{1+z^3}}$ (here $U(a)$ denotes some neighbourhood of point $a.$)

Let $f(\zeta)=\ln\frac{1-\zeta^3}{1+\zeta^3}, \;\; \zeta\in U(0).$ Then \begin{gather} \dfrac{df}{d\zeta}=-\dfrac{6\zeta^2}{1-\zeta^{6}}=-{6}\zeta^2 \sum\limits_{k=0}^{\infty}(\zeta^6)^k=-{6} \sum\limits_{k=0}^{\infty}\zeta^{6{k}+2}. \tag{*} \end{gather} Next you can integrate $(*)$ by $\zeta$ from $0\; \text{to}\; z.$