# Reference for zero sum estimates of Dirichlet L functions

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 $$\frac1{\rho (1-\rho)} = \frac1\rho + \frac1{1-\rho}.$$
• Why is $| \rho| \gg 1/ \log p$ ? Sorry I'm pretty new at this. Commented Sep 11, 2022 at 2:48
• The exceptional zero is by definition the only zero $\rho$ (if one exists) that does not satisfy $$|1-\rho| \gg 1 / \log p.$$ More fully: it is known that there exists $c>0$ such that all zeros satisfy $|1-\rho| > c / \log p$ with at most one exception. Commented Sep 11, 2022 at 14:30