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Q1. How does $F(\zeta) = F(z) = 0 \implies F'(z) = 0$? What about the denominator? Don't we have a $0/0$ situation?

Q2. How does continuity imply $F$ and its derivatives vanish at $w$?

Lemon
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1 Answers1

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1) Because the function $G(\zeta):=\dfrac{F(\zeta)-F(z)}{\zeta-z}$ on $\mathcal N\backslash\{z\}$ is the zero function, i.e, $G(\zeta)=0$ for all $\zeta\in\mathcal N\backslash\{z\}$, therefore its limit for $\zeta\to z$ is $0$.

We're not in the $0/0$ situation. It's as asking if $$\lim_{x\to 0}\dfrac{0}{x}=0$$ is wrong because it's the $0/0$ situation.

2) Let $F^{(n)}$ be the $n$-th derivative of $F$ for some $n\in\mathbb N$ (with $F^{(0)}=F$ by convention). Since $F$ is analytical on $D$, $F^{(n)}$ is continuous on $D$. In particular, for any parametric curve $\gamma(t)$ in $D$ so that $\gamma(1)=w$, we have $\lim_{t\to 1^-}F^{(n)}(\gamma(t))=F^{(n)}(w)$. In particular for the curve used in the proof, we have $F^{(n)}(\gamma(t))=0$ for $t\in (0,1)$ (so that $\gamma(t)$ is inside $\mathcal N$ for $t\in (0,1)$. Thus $F^{(n)}(w)=0$.

Scientifica
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  • How would that contradict statement 2? All we have shown is that only some points near $w$ are $0.$ It doesn't show that every point near $w$ is near – Lemon May 17 '20 at 04:14
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    @Hawk It's what's said in the next page, namely: since for all $n\in\mathbb N$, $F^{(n)}(w)=0$, the Taylor series of F at $w$ is $0$, i.e: since $F$ is analytical, there is a disk $B$ centered at $w$ such that $$\forall x\in B,,F(x)=\sum_{n=0}^\infty \dfrac{F^{(n)}(w)}{n!}(x-w)^n=0.$$ This contradicts statement (ii) because if there are points $z$ on $\Gamma$ arbitrary close to $w$ such that $F(z)\neq 0$, then we can find $z\in\Gamma\cap B$ such that $F(z)\neq0$. – Scientifica May 17 '20 at 09:33
  • I see because the convergence is a disk, it must touch those points that are not $0$. I went away for a few days to ponder about this result. Is there a reason that it fails in say $\mathbb{R}^2$? – Lemon May 20 '20 at 02:11
  • In $\mathbb R^2$ the notion of derivative is quite different than $\mathbb C$ to begin with, because we can't divide by vectors of $\mathbb R^2$ unlike in $\mathbb C$ where division by complex number makes sense. With the correct definition of differentiability for such a function, the result would actually hold. See this link. – Scientifica May 20 '20 at 04:05