I am trying to study the proof to the following theorem: If $|t| \geq 7/8$ and $5/6 \leq \sigma \leq 2$, then $$ \frac{\zeta'}{\zeta}(s) = \sum_{\rho}\frac{1}{s-\rho} + O({\log \tau})$$ where $\tau = |t| + 4$ and the sum is extended over all zeros $\rho$ of $\zeta(s)$ for which $|\rho - (3/2 + it)| \leq 5/6$
Using the lemma: Suppose that $f(z)$ is analytic in a domain containing the disc $|z| \leq 1$, that $|f(z)| \leq M$ in this disc, and that $f(0) \neq 0$. Let $r$ and $R$ be fixed, $0 < r < R < 1$. Then for $|z| \leq r$ we have $$ \frac{f'}{f}(z) = \sum_{k=1}^K \frac{1}{z-z_k}+ O({\log\frac{M}{|f(0)|}}) $$ where the sum is extended over all zeros $z_k$ of f for which $|z_k| \leq R$.
And
Let $\delta > 0$ be fixed. Then $ \zeta (s) = \frac{1}{s - 1} + O(1)$ uniformly for s in the rectangle $\delta \leq \sigma \leq 2$, $|t| \leq 1$, and $\zeta (s) \ll (1 + \tau^{1-\sigma} ) \min\{|\sigma - 1|,\log \tau\}$ , uniformly for $\delta \leq \sigma \leq 2, |t| \geq 1.$
And the the proof of the theorem given in the book is: We apply lemma 1 to the function $f (z) = \zeta (z + (3/2 + it))$, with $R = 5/6$ and $r = 2/3$. To complete the proof it suffices to note that $| f (0)|\gg 1$ by the (absolutely convergent) Euler product formula, and that $f (z) \ll \tau$ for $|z| \leq 1$ by Corollary lemma 2
My question is that when does the author use the assumption $|t|\geq 7/8$ or where is it implied?
The book I am referring is MULTIPLICATIVE NUMBER THEORY I:CLASSICAL THEORY by HUGH L. MONTGOMERY and ROBERT C. VAUGHAN
Edit: Thank you @Gary for pointing out the error of me mentioning the wrong authors of the book Edit: The theorem can be found in chapter 6 lemma 6.4 pg no.171. of the book mentioned above