I've been trying to learn some complex analysis on my own and I've got stuck on the following problem:
Let $f$ be continuous on the closed unit disc and holomorphic on the open unit disc. Let $n,k \in \mathbb{N}$ and $z_{nk}=\big(1-\frac{1}{n}\big)e^{i\frac{2\pi k}{n}}$ such that $f(z_{nk})=0.$ Prove that $f$ is identically zero.
The first thing that came to my mind was to use the Identity Theorem but it doesn't work here as $z_{nk}$ tends to $1$ as $n$ tends to $\infty$ and $1$ is on the boundary of the unit disc.
My idea is that I should prove that $f$ vanishes on the whole boundary of the unit disc because then I could apply the maximum principle and immediately conclude.
Any help would be much appreciated! Thanks!
Best Answer
I found this problem interesting so I wrote a bit more detailed answer, hope you enjoy it.
As you say, you need to prove $\forall z\in U:f(z) = 0$, from there you say you know how to conclude the final goal. Let's fix such a $z \in U$.
$f$ is continuous at $z$. Also, we have "a lot" of points $z_{nk}$ close to $z$ where $f$ is zero. If we construct a sequence of such points $\{w_i = z_{nk}\,\exists n,k\}_{i=0}^{\infty}$ that converges to $z$, this will mean $\{f(w_i) = 0\}_{i=0}^{\infty}$ converges to $f(z)$, hence $f(z) = 0$.
The rest of the proof is to build such a sequence from the numbers $z_{kn}$.
We have a $z$, let's fix some $\varepsilon >0$ and find a $z_{nk} \in B_{\varepsilon}(z)$.
Thanks to @WoolierThanThou who commented how to find one!
I'm really tempted to draw a picture to show the visual aspect of the proof:
Observe that every $z_{nk}$ lies on some line which joins $\{z|\text{Im}(z)=0\}$ at a rational angle. We'll fix such a line which passes through $B_{\varepsilon}(z)$ and look for a point $z_{nk}$ on it which lies inside that $B_{\varepsilon}(z)$. On my picture, the line is $l$, and we're looking for a $z_{kn}$ on the bolded segment, denote its length by $\varepsilon'$.
I'll leave the existence of a $t \in \mathbb Q\cap(0,1)$ such that $l = \{ r e^{t 2\pi i} | r \in \mathbb R \}$ contains an inner point of $B_{\varepsilon}(z)$ for you. Hint: look at the interval $(\text{Arg}(a), \text{Arg}(b))$ where $\{ a, b \} = U \cap \overline{B_{\varepsilon}(z)}$.
Let's take such a $t$ and focus on the set (the bold segment on my picture): $$l \cap B_{\varepsilon}(z) = \{re^{t2\pi i}| r \in \mathbb R \cap (1-\varepsilon', 1) \} = \{(1-r)e^{t2\pi i} | 0 < r < \varepsilon' \}$$
Express $t$ in the form $k / n$ where $k,n\in\mathbb N$ so that $n$ is sufficiently large: $n > 1/\varepsilon'$. Then clearly $(1-1/n)e^{t2\pi i}$ lies in this set (take $r = 1/n < \varepsilon'$), and that's in fact the number $z_{nk} = (1-1/n)e^{\frac{k}{n}2\pi i}$.
Done - we found a $z_{nk}$ lying in $B_{\varepsilon}(z)$ for any $\varepsilon>0$ and any $z\in U$. This means, there's a converging sequence of $z_{nk}$'s to any $z\in U$ and so $f(z) = 0$ for all $z \in U$.