Let $$g:B_1(1):=\left\{z\in\mathbb{C} :|z-1|<1\right\}\to\mathbb{C}\;,\;\;\;z\mapsto\ln z-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(z-1)^n$$ (1) In a first step, I'm asked to show, that $g$ is sonstant, i.e. $\text{im }g=\left\{c\right\}$ for some $c\in\mathbb{C}$. I think I would need to:
- show that the power series converges for all $z\in B_1(1)$
- use the fact, that each power series is complex differentiable inside its circle of convergence
- use the fact, that $(\ln z)'=z^{-1}$ for all $z\in B_1(1)$
- show that $$\left(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(z-1)^n\right)'=\sum_{n=0}^\infty (1-z)^n=\frac{1}{z}$$ for all $z\in\mathbb{C}$
It follows $$g'(z)=\frac{1}{z}-\frac{1}{z}=0$$
(2) From (1) it follows $$\ln z=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(z-1)^n+c$$ for some $c\in\mathbb{C}$. What's the reason why it must hold $c=0$?