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\begin{fclemma}[Landau]
  \label{lemma:landau}
  Assuming:
  - $z_0 \in \Cbb$ and $r > 0$
  - $f$ analytic in $B(z_0, r)$
  - for some $M > 1$ we have $|f(z)| < e^M
  |f(z_0)|$ for all $z \in B(z_0, r)$
  Then:
  for $z \in B(z_0, r / 4)$,
  \[
    \left| \frac{f'(z)}{f(z)} - \sum_{\rho \in Z}
    \frac{1}{z - \rho} \right|
    \le \frac{96 M}{r}
  ,\]
  where $Z$ is the set of zeroes of $f$ in
  $\ol{B(z_0, r / 2)}$, counted with
  multiplicities.
\end{fclemma}

\begin{note*}
  If $f$ is a polynomial, we can factorise
  \[
    f(z) = a \prod_\rho (z - \rho)
  ,\]
  and then
  \begin{align*}
    \frac{f'(z)}{f(z)}
    &= (\log f(z))' \\
    &= \left( \log a + \sum_\rho \log(z - \rho)
    \right)' \\
    &= \sum_\rho \frac{1}{z - \rho}
  \end{align*}
\end{note*}

\begin{proof}
  Let
  \[
    g(z) = \frac{f(z)}{\prod_{\rho \in Z} (z -
    \rho)}
  .\]
  Then $g$ is analytic and non-vanishing in
  $\ol{B(z_0, r/2)}$. Note that
  \[
    \frac{g'(z)}{g(z)}
    = \frac{f'(z)}{f(z)}
    - \sum_{\rho \in Z} \frac{1}{z - \rho}
  \]
  ($\frac{(f_1 - f_n)'}{f_1 - f_n} = \sum_{i =
  1}^n \frac{f_1'}{f_1}$).

  Hence, it suffices to prove
  \[
    \left| \frac{g'(z)}{g(z)} \right|
    \le \frac{96 M}{r}
  \]
  for $z \in B(z_0, r / 4)$. Write
  \[
    h(z) = \frac{g(z_0 + z)}{g(z_0)}
  .\]
  Then $h$ is analytic and non-vanishing in
  $\ol{B(0, r / 2)}$, and $h(0) = 1$. We want to
  show
  \[
    \left| \frac{h'(z)}{h(z)} \right|
    \le \frac{96M}{r}
  \]
  for $z \in B(0, r / 4)$.

  For all $z_0 \in \partial B(0, r)$ we have
  \[
    |h(z)|
    = \left| \frac{f(z_0 + z)}{f(z_0)} \prod_{\rho
    \in Z} \frac{z_0 - \rho}{z_0 + z - \rho} \right|
    \le \left| \frac{f(z_0 + z)}{f(z_0)} \right|
    < e^M
  \]
  since
  \[
    |z_0 - \rho|
    \le \frac{r}{2}
    = r - \frac{r}{2}
    \le |z_0 + z - \rho|
  \]
  for $z \in \partial B(0, r)$.

  By the maximum modulus principle, $|h(z)| < e^M$
  for $z \in \ol{B(0, r / 2)}$, so
  \[
    \Re \log h(z)
    = \log |h(z)|
    < M
  .\]
  By the \nameref{thm:borel-cara} with radii
  $\frac{3r}{8}$, $\frac{r}{4}$ we have for for $z
  \in B(0, 3r/8)$,
  \[
    |\log h(z)|
    \le \frac{2 \frac{r}{4}}{\frac{3r}{8} -
    \frac{r}{4}} M
    = 4M
  .\]
  Now, for $z \in B(0, r / 4)$, Cauchy's theorem
  gives us
  \begin{align*}
    \left| \frac{h'(z)}{h(z)} \right|
    &= \left| \frac{1}{2\pi i} \int_{\partial B(0,
    3r/8)}  \frac{\log h(w)}{(z - w)^2} \dd w
    \right| \\
    &\le \frac{1}{2\pi} \cdot 2\pi \frac{3r}{8} 4M
    \left( \frac{3r}{8} - \frac{r}{4} \right)^{-2}
    \\
    &= \frac{96M}{r}
    \qedhere
  \end{align*}
\end{proof}

\begin{fcthm}[Partial Fraction approximation of $zeta' / zeta$]
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $s = \sigma + it$ with $|\sigma| \le
      10$, $s \neq 1$ and $\zeta(s) \neq 0$. Then
      \[
        \frac{\zeta'(s)}{\zeta(s)}
        = -\frac{1}{s - 1}
        + \sum_{|\rho - s| \le \frac{1}{10}}
        \frac{1}{s - \rho} + O(\log(|t| + 2))
      .\]
      where the sum is over the zeroes $\rho$ of
      $\zeta$ counted with multiplicity.
    \item For any $T \ge 0$, there are
      $\vll \log (T + 2)$ many zeroes $\rho$ of
      $\zeta$ (counted with multiplicity) with
      $|\Im(\rho)| \in [T, T + 1]$.
  \end{enumerate}
\end{fcthm}

\begin{proof}
  We apply \nameref{lemma:landau} with $z_0 = 2 +
  it$, $r = 50$, with $f(s) = (s - 1) \zeta(s)$.

  By the lemma on polynomial growth of $\zeta$ on
  vertical lines, for $\sigma + it \in B(z_0,
  50)$, we have
  \begin{align*}
    |f(s)|
    &\le C(|t| + 52)(|t\ + 2)^{50} \\
    &\le C \exp(51 \log (|t| + 52)) \\
    &\vll C \exp(51 \log (|t| + 52)) |f(z)|
  \end{align*}
  ($|\zeta(z + it)| \asymp 1$).

  Let $s \in B \left( z_0, \frac{25}{2} \right)$.
  Then \nameref{lemma:landau} gives us
  \begin{align*}
    \frac{f'(s)}{f(s)}
    &= \frac{1}{s - 1} +
    \frac{\zeta'(s)}{\zeta(s)} \\
    &= \sum_{|\rho - z_0| \le 2S} \frac{1}{s -
    \rho} + O (\log(|t| + 2))
    \tag{$*$}
    \label{lec15eq1}
  \end{align*}
  Since $B \left( z_0, \frac{25}{2} \right)$
  contains all the points $s = \sigma + it$ with
  $|\sigma| \le 10$, it suffices to show
  \[
    \sum_{\substack{|\rho - z_0| \le 25 \\ |\rho -
    s| > \frac{1}{10}}} \frac{1}{s - \rho}
    = O(\log(|t| + 2))
    \tag{$**$}
    \label{lec15eq2}
  .\]
  Substituting $s = z_0$ in \eqref{lec15eq1}, we
  get
  \begin{align*}
    \sum_{|\rho - z_0| \le 25}
    \frac{1}{z_0 - \rho}
    &= O \left( \left|
    \frac{\zeta'(z_0)}{\zeta(z_0)} \right| + 1 +
    \log(|t| + 2) \right) \\
    &= O(\log(|t| + 2))
  \end{align*}
  since
  \begin{align*}
    \left| \frac{\zeta'(2 + it)}{\zeta(2 + it)} \right|
    &= \left| \sum_{n = 1}^\infty \vman(n) n^{-2 +
    it} \right| \\
    &\le \sum_{n = 1}^\infty \vman(n) n^{-2} \\
    &= -\frac{\zeta(2)}{\zeta(2)}
  \end{align*}
  Taking real parts,
  \[
    \sum_{|\rho - z_0| \le 25} \frac{2 -
    \Re(\rho)}{|z_0 - \rho|^2}
    = O(\log(|t| + 2))
  .\]
  Since $\Re(\rho) \le 1$,
  \[
    \sum_{|\rho - z_0| \le 25} \frac{1}{|z_0 -
    \rho|^2}
    = O(\log(|t| + 2))
  .\]
  This proves part (ii). It gives also
  \eqref{lec15eq2} since the sum there contains
  $O(\log(|t| + 2))$ zeros.
\end{proof}