If $\rho=\beta+i\gamma$, $\prod_\rho\left(1-\frac s\rho\right)=\prod_{\gamma>0}\left(1-\frac{s(1-s)}{\rho(1-\rho)}\right)$

Question

Next we consider the convergence of the product. With $\rho=\beta+i\gamma$ let $$\prod_\rho\left(1-\frac s\rho\right)=\prod_{\gamma>0}\left(1-\frac{s(1-s)}{\rho(1-\rho)}\right).$$ This converges absolutely provided $\sum_{\gamma>0}1/(\rho(1-\rho))<\infty$. To see this, complete the square.

How do we get the above equality. I know that if $\rho$ is a zero then $1-\rho$ is a zero of Riemann's xi function. But how do we get $\operatorname{Im}\rho=\gamma>0$?