Complex analysis: integrating a power series term-by-term

Question

If $f(z)=\sum_{n=0}^{\infty}a_nz^n,\,(a_n\in\mathbb{C})$ for each $z\in B(0,R)$, then $$ \int_0^z f(\zeta)d\zeta=\sum_{n=0}^{\infty}a_n\frac{z^{n+1}}{n+1},\,\forall z\in B(0,R).$$ The integration happens over a contour $\Gamma\subseteq B(0,R)$.

Given proof:

Choose a contour $\Gamma\subseteq B(0,R)$ from $0$ to $z$. Contour $\Gamma$ is compact, so $d(\Gamma,\mathbb{C}\backslash B(0,R))>0$, which implies that $\Gamma \subseteq B(0,r)$ for some $r<R$. ...

I don't see where the last implication comes from. Does it have to do with the fact that $B(0,R)$ is open and therefore $\forall z\in B(0,R), \exists \varepsilon >0: B(z,\varepsilon)\subseteq B(0,R)$. Since $z$ is the end point of $\Gamma$, we can choose $r := R-\varepsilon$. Is this correct?

Thanks.

Answer 1

Consider the set$$\left\{B\left(0,R-\frac Rn\right)\,\middle|\,n\in\mathbb N\setminus\{1\}\right\}.$$It is an open cover of $\Gamma$, which is compact. Therefore, it has a finite subcover$$\left\{B\left(0,R-\frac Rn\right)\,\middle|\,n\in\{n_1,n_2,\ldots,n_k\}\right\}.$$Now, let $N=\max\{n_1,n_2,\ldots,n_k\}$. Then $\Gamma\subset B\left(0,R-\frac RN\right)$. And $R-\frac RN<R$.

Answer 2

Let $r=R-d(\Gamma,\mathbb{C}\backslash B(0,R))<R$. For any $0\neq z\in\Gamma\subset B(0,R)$, we have $$d(z,\mathbb{C}\backslash B(0,R))\geq d(\Gamma,\mathbb{C}\backslash B(0,R))=R-r.$$ Suppose that the line $Oz$ intersects $B(0,R)$ at $z'$, then $$d(z,z')=d(z,\mathbb{C}\backslash B(0,R))=R-r,$$ and then $d(z,O)=d(z',O)-d(z,z')\leq R-(R-r)=r$ since $z$ is between $O$ and $z'$.