I'm going through my lecture notes and I'm stuck on the proof of
For any $t>0$ and primitive $\chi$ modulo $q$
$$\sum_{\rho=\beta+i \gamma: \Lambda(\rho, \chi)=0}\frac{1}{1+(t-\gamma)^2}=O(\log q+\log(|t|+2))$$
where $\Lambda(s, \chi)=\left(\frac{q}{\pi}\right)^{\frac{1}{2}(s+a)}\Gamma\left(\frac{1}{2}(s+a)\right)L(s, \chi)$ is the completed $L$-function, with $\chi$ primitive and $a=0$ if $\chi$ is even, $a=1$ if $\chi$ is odd.
The lecturer writes
Proof. Write $s=\sigma+it$, $\rho=\beta+i\gamma$. For $|\sigma|$ bounded,
$$\frac{-L'}{L}(s, \chi)=O(\log(|t|+2)+\log q)-B(\chi)-\sum_{\rho} \left( \frac{1}{s-\rho}+\frac{1}{\rho}\right)$$
Now, I'm not convinced this is true without $|\sigma+a|$ also being bounded below since we have
$$\frac{-L'}{L}(s, \chi)=\frac{1}{2}\log \frac{q}{\pi}+\frac{\Gamma'}{\Gamma}\left(\frac{s+a}{2}\right)-B(\chi)-\sum_{\rho} \left( \frac{1}{s-\rho}+\frac{1}{\rho}\right).$$
The $\frac{1}{2}\log \frac{q}{\pi}$ gets dealt with by $O(\log q)$.
Stirlings approximation gives $$\frac{\Gamma'}{\Gamma}\left(\frac{s+a}{2}\right)=\log\left(\frac{s+a}{2}\right)+O\left(\frac{2}{s+a}\right) $$
and $O\left(\frac{2}{s+a}\right)$ gets big if we let $s+a$ get small. Another problem is we can only use Stirling's and get a uniform bound if $\sigma>0$, and I think taking $\sigma>0$ would also give us a proof for all $t \in \mathbb{R}$.
This kinda doesn't matter immediately, because in the next line we let $\sigma=2$, and it's definitely true that
$$\frac{-L'}{L}(2+it, \chi)=O(\log(|t|+2)+\log q)-B(\chi)-\sum_{\rho} \left( \frac{1}{2+it-\rho}+\frac{1}{\rho}\right).$$
The rest of the proof goes
Now we let $\sigma=2$ and take real parts to deduce $$\mathfrak{R}\frac{-L'}{L}(2+it)=O(1)=O(\log(|t|+2)+\log q)-\sum_\rho \frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2}.$$
This proves the first set of statements. In treating $\zeta$, the above formulae contains in addition the terms $\frac{1}{s}+\frac{1}{s-1}$ which are $O(1/|t|)$.
I don't see this statement about $\zeta$ either. We have
$$O(1)=\frac{-\zeta'}{\zeta}(2+it)=-\frac{1}{2}\log \pi+\frac{\Gamma'}{\Gamma}\left(\frac{2+it}{2}\right)+\frac{1}{2+it}+\frac{1}{1+it}-B-\sum_{\rho} \left( \frac{1}{2+it-\rho}+\frac{1}{\rho}\right)$$
and $\frac{1}{2+it}+\frac{1}{1+it}=O(1)$, so don't we get exactly the same bound?