I have to prove that $f(Z)=\log((1+z)/(1-z))$,$z\in \mathbb{C}$ it can expand a series of taylor like $f(z)=2\sum_{n=0}^{\infty} ( x^{2n+1})/(2n+1)$ for all $z\in D(0,1)$
my idea is to drift multiple times and find the recursion:
$f^{(1)}(z)=\frac{2(1-z)}{(1-z)^2(1+z)}$
$f^{(2)}(z)=\frac{z^2+1}{(z+1)^2(z-1)^2}$
$f^{(3)}(z)=\frac{2(-z^2+z-2)}{(1+z)^5}$
$f^{(4)}(z)=\frac{2(3z^2-6z+11)}{(1+z)^6}$
I need to find the recursion to test this but can't find generalize this.
On the other hand, Taking into account that $\boxed{\log(1-z) = -\sum_{n = 1}^{+\infty} \frac{z^n}{n}}$ and $\log(1+z) = \sum_{n= 1}^{+\infty} (-1)^{n+1} \frac{z^n}{n}$
but I don't get what I want using this
Hint:- Even if the above comment is sufficient (and definitely easier)you could try the euler representation of the z since its within the range of (0,1). Then proceed to use half angle trigonometric relations such as
$1+cosx=2cos^2(x/2)$
$1-cosx=2sin^2(x/2)$
$sinx=2sinx/2.cosx/2$