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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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    Both approaches should give you the same answer.2012-12-12
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    Are you looking for the complete series, or just the first few terms? The answer will affect the preferred approach.2012-12-12
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    @Mhenni Benghorbal: Is the first approach legal in the complex case?2012-12-12
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    @Mario Carneiro: I'm looking for the complete series.2012-12-12
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    You can still use the identity $\log\frac ab=\log a-\log b$ in the complex case, but you have to be careful about the branch cuts. If you are plugging it into a computer, it may not hold, but on paper, you are just saying that the set of values taken on the left hand side is the same as the set on the right, which is still true.2012-12-12
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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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