Find the spectrum of the linear operator $T: \ell^2 \to \ell^2$ defined by $Tx=(\theta x_{n-1} +(1-\theta)x_{n+1})_{n\in \mathbb{Z}}$

To determine the spectrum of $T$, let us first determine the one of the right shift $\tau$. Since $||\tau||=1$, $\mathrm{Sp}(\tau) \subset \bar{B}(0,1)$. But the same goes for the left shift $\tau^{-1}$, so $\mathrm{Sp}(\tau) \subset C(0,1)$. It is actually equal to $C(0,1)$: $(\ldots,0,1,\lambda,\lambda^2,\ldots,\lambda^n,0,\ldots)$ is an "almost eigenvector".

For every $c \in \mathbb{C}^{\times}$, $(c \theta \mathrm{Id} - \tau)(c (1-\theta) \mathrm{Id} - \tau^{-1}) = (1+c^2 \theta (1-\theta)) \mathrm{Id} - cT$, so $f(c)=\frac{1+c^2 \theta (1-\theta)}{c} \in \mathrm{Sp} (T)$ iff $c \theta \in \mathrm{Sp}(\tau)$ or $c (1- \theta) \in \mathrm{Sp}(\tau^{-1})$ iff $|c|=\theta^{-1}$ or $(1-\theta)^{-1}$.

Now $f(\mathbb{C}^{\times})=\mathbb{C}$, and $f(\theta^{-1} (1-\theta)^{-1} c^{-1})=f(c)$, so $\mathrm{Sp}(T)= \left\{ f(\theta e^{i \alpha}) \right\} = \left\{ \cos \alpha + i (1-2 \theta) \sin \alpha \right\}$ which is an ellipsis (flat when $\theta=1/2$).

EDIT: It is easy to check that for all $\lambda$, $\lambda \mathrm{Id} - \tau$ is injective and has dense range, hence the same is true for $T$.