When trying to find the numerical range $W(T) := \{ \langle Tx, x \rangle: \| x \| = 1\}$ of the left shift operator $T_2: \ell_2 \to \ell_2, \ (x_1, x_2, \ldots) \mapsto (x_2, \ldots)$ I found it to be $$W(T_2) = \left\{ \sum_{k = 1}^{\infty} x_{k} \overline{x_{k + 1}} : \sum_{k = 1}^{\infty} | x_k |^2 = 1\right \}, $$ as $x_k$ can be complex, too.
This seems to be closely related to this question of mine, where I wanted to characterise $\{ \overline{x_1} x_2: \| x \| = 1\}$. This is relatively easy to argue with the polar representation of $x_1$ and $x_2$, as can be seen in the comments of the question.
How do I find this set? Is there a similar approach?
First of all, note that $x\mapsto \langle Tx,x\rangle$ is $\ell^2$-continuous and maps $(1,0,0,...)$ to $0$. Furthermore, note that for any $\theta\in \mathbb{R},$ we have that if $||x||_2=1$, then the sequence $x_{\theta}$ given by $(e^{i\theta}x_1,x_2,x_3,...),$ we see that $\langle Tx_{\theta},x_{\theta}\rangle= e^{i\theta}\langle Tx,x\rangle.$ Thus, $e^{i\theta} W(T)=W(T)$. Adding this together implies that $W(T)$ is some disk.
Now, let $x^k=(x^k_n)_{n\in \mathbb{N}}$ be given by $x^k_n=\frac{1}{\sqrt{k}}$ for $n\leq k$ and $x^k_n=0$ else. Then, we see that $||x^k||_2=\sqrt{k\frac{1}{k}}=1$ for each $k$. Furthermore,
$$ \langle Tx^k, x^k\rangle=\frac{k-1}{k} $$ As this tends to $1$ for $k\to \infty,$ we get that $\mathbb{D}\subseteq W(T)$, where $\mathbb{D}$ denotes the open unit disk. Furthermore, by Cauchy-Schwarz, for any $\ell^2$-sequence $x$,
$$ |\langle Tx,x\rangle|\leq ||Tx||_2||x||_2\leq ||x||_2^2, $$
where we have used that $T$ is a contraction. This implies that $W(T)$ is contained in the closed unit disk. Since $W(T)$ is a disk, it suffices to argue that $1\not \in W(T)$.
Assume some sequence $x$ such that $\langle Tx,x\rangle=||x||_2^2$. Then, by the above argument, we must have $||Tx||_2=||x||_2$. This clearly implies that $x_1=0$. However, we also have equality in the Cauchy-Schwarz inequality, but this happens if and only if there exists some $\lambda\in \mathbb{C}$ such that $Tx=\lambda x$. Then, $x_2=\lambda x_1=0$. Inductively, we get that $x$ is the $0$ sequence.
In conclusion, we find that $1\not\in W(T)$ and hence, $W(T)=\mathbb{D}$.