Let $E$ be a complex Hilbert space. Let $A\in \mathcal{L}(E)$. Consider \begin{eqnarray*} W_{0}(A) &=&\{\alpha\in \mathbb{C}:\;\exists\,(z_n)\subset E\;\;\hbox{such that}\;\|z_n\|=1,\displaystyle\lim_{n\rightarrow+\infty}\langle A z_n,z_n\rangle=\alpha,\\ &&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}\|Az_n\|= \|A\| \}. \end{eqnarray*}
If $0\notin W_{0}(A)$, by the transformation $Ae^{i\alpha}$ of $A$, why we can suppose that $\Re e( W_{0}(A))\geq\tau>0$. Notice that this result figures in the proof of THEOREM $2$. (1)
Invariance of the inequality to a rotation
By assumption we have that, $$ \| T \| \leq \| T + \lambda \| \hspace{1cm} \forall \lambda\in \mathbb{C} $$
Let $\lambda = \alpha\beta $, where $|\alpha| = 1$, it follows that, $$ \begin{align} \| T + \lambda \| =& \| T + \alpha\beta \| \\ =& \| \bar{\alpha} T + \beta \| \\ \geq& \| \bar{\alpha} T \| = \|T\| \end{align} $$
Which implies that the inequality is invariant to arbitrary rotation.
Expanding the vector norm to reveal the inner product
By definition we have that, $$ \begin{align} \| T + \lambda \|^2 =& \sup_{\|x\|=1} \|(T + \lambda)x\|^2 \end{align} $$ also, $$ \begin{align} \|(T + \lambda)x\|^2 =& \langle(T + \lambda)x, (T + \lambda)x \rangle \\ =& \|Tx\|^2 + \lambda^2 + \langle Tx, x\rangle+ \langle x, Tx\rangle \\ =& \|Tx\|^2 + \lambda^2 + 2 \mathcal{Re}\langle Tx, x\rangle \end{align} $$ Suppose that $\alpha \in \mathbb{C}$ is a rotation, as before, then, $$ \begin{align} \|(T + \lambda)x\|^2 =& \|Tx\|^2 + \beta^2 + 2 \mathcal{Re}\left(\alpha\langle Tx, x\rangle\right) \end{align} $$ and so we can rotate the complex number $\alpha\langle Tx, x\rangle$ so that it has positive real part, without affecting the assumed inequality.
Equivalence of $W_0(T)$ to a rotated set
Suppose that we have a sequence $z_n \in H $, for which $\| Tz_n \| \to \|T\|$ and $\langle Tz_n, z_n \rangle \to \rho $. Using the expansion in the section above gives, $$ \begin{align} \|(T + \lambda)z_n\|^2 =& \|Tz_n\|^2 + \lambda^2 + 2 \mathcal{Re}\left(\langle Tz_n, z_n\rangle\right) \\ \to& \|T\| + \lambda^2 + 2\rho \hspace{ 1cm } \mbox{ as } n \to \infty \end{align} $$
If we now choose the sequence $\alpha_n$, $\beta_n$ that ensures that $\mathcal{Re}(\alpha\langle Tz_n, z_n \rangle) > 0$, for each $n$, we have that, $$ \begin{align} \|(T + \lambda)z_n\|^2 =& \|Tz_n\|^2 + \beta_n^2 + 2 \mathcal{Re}\left(\alpha_n\langle Tz_n, z_n\rangle\right) \\ \to& \|T\| + \lambda^2 + 2\rho \hspace{ 1cm } \mbox{ as } n \to \infty \end{align} $$
It turns out that the $ \xi_n = \alpha_n\langle Tz_n, z_n\rangle $ are bounded, since $T$ is a bounded linear operator, which means that there is a convergent subsequence $\xi_{n_k}$ for which $\|(T + \lambda)z_{n_k}\|^2$ reaches the same limit.
So now for each sequence $z_n \in H$ there is an alternative sequence $y_k = \xi_{n_k}$ with $\|y_k\|=1$ for which $ \alpha_k\langle Ty_k, y_k\rangle \to \rho$ and $\|Ty_k\| \to \|T\|$ as $k\to\infty$.
This means that, $$ W_0(T) = \{ \rho : \exists z_n \mbox{ with } \|z_n\|=1 \mbox{ , } \alpha_n \langle Tz_n, z_n \rangle \to \rho \mbox{ and } \|Tz_n\| \to \|T\| \} $$ where $\alpha_n$ has been chosen so that $\mathcal{Re}(\alpha_n \langle Tz_n, z_n \rangle) > 0$.
Every element of $\mathcal{Re}(W_0(T))$ must be greater than zero
Now we have that $\mathcal{Re}(\alpha_n \langle Tz_n, z_n \rangle) > 0$ for each $n$, and also, $\mathcal{Re}(\alpha_n \langle Tz_n, z_n \rangle) \to \mathcal{Re}(\rho)$ as $n \to \infty$. Since $0 \notin W_0(T)$ it follows that $\rho \neq 0$ and so there exists a $\tau > 0$ such that $\mathcal{Re}(W_0(T)) \geq \tau > 0$.