Let $f:\{z:\|z\|<1\}\rightarrow \{z:-\frac{\pi}4<\operatorname{Im}(z)<\frac {\pi}4\}$ such that $f (z)=\sum\limits_{n=1}^\infty\dfrac{z^{2n-1}}{2n-1}$. How to prove that $f(z)$ is analytic and also injetive. I have been able to show that $f$ is analytical due to the analytic convergence theorem:
If $(f_k)_{k\in\mathbb{N}}$ sequence of analytic functions defined in a region $A\subseteq\mathbb{C}$ and $f(z)=\displaystyle\sum_{k=1}^\infty f_k(z)$ uniformly converges in any closed disk in $A$, then $f(z)$ is analytic in $A.$
But I have not been able to demonstrate that $f(z)$ is injective, could you give me any idea about it?
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Meaning the principal branch of log, we can sum the series (it's sometimes known as the inverse hyperbolic tangent): $$ 2f(z)=2\sum_{n=0}^{\infty}\frac{z^{2n+1}}{2n+1} = \log(1+z)-\log(1-z)=\log\left( \frac{1+z}{1-z}\right) $$If $f(z)=f(w)$, then $$ \log\left( \frac{1+z}{1-z}\right)=\log\left( \frac{1+w}{1-w}\right) $$Exponentiating, we see that $$ -1 - \frac{2}{1-z}=-1 - \frac{2}{1-w}, $$or $z=w$.