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**Drexel28** $\displaystyle \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)$- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.

All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.

So $\displaystyle \sum_{\rho,\text{ }\Im \rho}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)$

But clearly this is equal to

$\displaystyle \sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im\rho>0}f(\overline{\rho})$- We just changed in the second sum where the conjugation occurs.

But, we can add these two together to get $\displaystyle \sum_{\rho,\text{ }\Im \rho>0}\left\{f(\rho)+f(\overline{\rho})\right\}$.

This is assuming everything converged.