Question Prove that Riemann Hypothesis is equivalent to the statement that $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$ where $\rho$ are the non trivial zeros of Riemann zeta function.
Attempt
Assume that RH is true.
Then, $\Re(\rho)=\frac{1}{2}$
Hence, $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Conversely Assume, $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Assume on the contrary that there exists some $\rho_0$ such that $\Re(\rho_0)\neq \frac{1}{2}$. Since $0<\Re(\rho)<1$, thus let, $0<\Re(\rho_0)<\frac{1}{2}$.
How to prove the converse part?.
Edit
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Since $\rho$ has $\Im(\rho)>0$ , so $1-\bar{\rho}$ has $\Im(1-\bar{\rho})>0$.
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2} + \sum_{\Im(\rho)>0}\frac{1-2\ \Re(1-\bar{\rho})}{|\frac{1}{2}-(1-\bar{\rho}|^2}
=0 $$
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2} + \sum_{\Im(\rho)>0}\frac{2\Re(\rho)-1}{|\frac{1}{2}-{\rho}|^2}
=0 $$
What to do next?
Best Answer
The claim in the OP is not true as $\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0$ unconditionally (the identity holds if RH true and if RH false)
By symmetry whenever $\rho$ is a root with $\Im \rho >0$, so is $1-\bar \rho$; but $\frac{1}{|1/2-\rho|^2}=\frac{1}{|\bar \rho -1/2|^2}$, while $\Re \rho + \Re (1-\bar \rho)=1$, so in the case of RH false we group the terms with $\Im \rho >0$ as above $\rho, 1-\bar \rho$ when they are distinct (in other the roots off the critical line with positive imaginary part) together and we get $0$, while the critical line zeroes give $0$ trivially as noted in the original post.