We want to show that for some holomorphic function $f:G\rightarrow \mathbb{C}$, $\forall z\in G$, and $|f(z)-1|<1$, $$\int_C\frac{f'(z)}{f(z)}dz=0$$
Where, $f'(z)$ is continuous on $G$ and $C$ is a closed curve in $G$.
This problem is easy because we just differentiate $\log(f(z))$ then use the closed curve theorem and the fundamental theorem of calculus for line integrals to deduce the result.
However what I fail to see is that for $\log(f(z))$ to be differentiable, it has to be continuous, and to show that it is continuous we have to show that $\arg(f(z))$ is continuous as clearly $\ln|f(z)|$ is continuous. This is the part I am stuck at. Could anyone help me so I can understand how $\log(f(z))$ is proven to be continuous?
It seems that you forgot that $(\forall z\in G):\bigl|f(z)-1\bigr|<1$. And in $D(1,1)$ you can define$$\log z=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n(z-1)^n,$$which is analytic and therefore ontinuous.