I'm studying about digital control systems and I was given this transformation $$s=\frac{2}{T}\frac{z-1}{z+1}$$ But I don't like just accepting things so I looked it up to find the proof and I found this on wikipedia 
I'm aware of the taylor and maclaurin series but I can't quite get what's happening in the second line. How is lnz equal to all that?
From the power series of the logarithm
$$\log(1+x) = - \sum_{n>0} \frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots$$
(when $\vert x \vert < 1)$, we can immediately deduce
$$\log(1-x) = - \sum_{n>0} \frac{x^n}{n}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\ldots,$$
and substracting the two expressions we can see that only the odd degree terms survive on the right hand side:
$$\log(1+x) - \log(1-x) = 2\sum_{n>0} \frac{x^{2n-1}}{2n-1}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right)$$
For the standard branch of the complex logarithm, we have the equality
$$\log(1+x) - \log(1-x) = \log\left(\frac{1+x}{1-x}\right).$$
Now substituting $z = \frac{1+x}{1-x}$, we get $x = \frac{z-1}{z+1}$, and we recover the expression in the second line of the image you linked.
