What method was used here to expand $\ln(z)$?

Question

On Wikipedia's entry for bilinear transform, there is this formula:

\begin{align} s &= \frac{1}{T} \ln(z) \\[6pt] &= \frac{2}{T} \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3 + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5 + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right] \\[6pt] &\approx \frac{2}{T} \frac{z - 1}{z + 1} \\[6pt] &= \frac{2}{T} \frac{1 - z^{-1}}{1 + z^{-1}} \end{align}

What is the method that is used to expand $\ln(z)$? Taylor series? Laurent series? Some other techniques?

Answer 1

Taylor series show $$\log((1-z)/(1+z)) = \log(1-z) - \log(1+z) = 2(z+z^3/3 + \cdots)$$ then let $y=(1-z)/(1+z)$.

The math works, though something doesn't jive with my intuition. I guess what's confusing is that the Taylor series already looks like log but it is missing the even terms. With that, as z goes to infinity the series approaches the divergent harmonic series, which makes sense.

Answer 2

$$ z = \frac{1+\frac{z-1}{z+1}}{1-\frac{z-1}{z+1}} $$