Let $\chi$ be a primitive character mod $p$ (prime) and $\rho = \beta + i \gamma$ be a non-trivial zero of $L(s, \chi)$. I am reading a paper by Ihara and Murty where they use following estimate :
$\left|\frac{1}{\rho(1-\rho)} \right| \ll \log p \;$ for $\; |\gamma| \leq 1$.
(We can assume $\rho$ is not the possible exceptional zero.) I am not sure why this is true. They just mention it is well-known, so any reference will also be helpful.
This follows from Noam's answer and the classical zero-free region for Dirichlet $L$-functions. For a specific reference, see Davenport's Multiplicative Number Theory, Chapter 14. The result quoted below can be found on page 93:
Theorem. There exists a positive absolute constant $c$ with the following property. If $\chi$ is a complex character modulo $q$, then $L(s,\chi)$ has no zero in the region defined by $$ \sigma \geq \begin{cases} 1 - \dfrac{c}{\log(q |t|)} & \text{if $|t| \geq 1$},\\[1em] 1- \dfrac{c}{\log(q)} & \text{if $|t| \leq 1$}. \end{cases} $$ Note that this holds for complex characters because in this case there is no exceptional zero. A similar result holds for quadratic characters, aside from the exceptional zero, of course.
We thus have, for any non-exceptional zero $\rho = \beta+i\gamma$ $$ |1-\rho| = |1-\beta - i\gamma| \geq 1-\beta \geq \frac{c}{\log(q)} $$ for $|\gamma|\leq 1$.
We must assume that $1-\rho$ is not exceptional either. Then both $|\rho|$ and $|1-\rho|$ are $\gg 1 / \log p$, and the desired inequality follows, for instance because $$ \frac1{\rho (1-\rho)} = \frac1\rho + \frac1{1-\rho}. $$
