nice complex exercise asked by Dan Stirua in many math groups

Exercise:

Show that , $\frac{\left| \sum\limits_{i=1}^{n}{\left( \operatorname{Re}\left( {{z}_{i}} \right)-\operatorname{Im}\left( {{z}_{i}} \right) \right)\times \sum\limits_{i=1}^{n}{\left( \operatorname{Re}\left( {{z}_{i}} \right)+\operatorname{Im}\left( {{z}_{i}} \right) \right)}} \right|}{{{\left( \sum\limits_{i=1}^{n}{\left| {{z}_{i}} \right|} \right)}^{2}}}\le 2$   ,   ${{\left\{ {{z}_{i}} \right\}}_{i\ge 1}}\in {{\mathbb{C}}^{n}}$

Solution : Let $z=a+ib\Leftrightarrow \bar{z}=a-ib$

So $z+\bar{z}=\left( a+ib \right)+\left( a-ib \right)=2a=2\operatorname{Re}\left( z \right)$ and $z-\bar{z}=a+ib-a+ib=2ib=2i\operatorname{Im}\left( z \right)$

Hence $\operatorname{Re}\left( z \right)=\frac{z+\bar{z}}{2}\,\,\,\,\And \,\,\operatorname{Im}\left( z \right)=\frac{z-\bar{z}}{2i}$

So $\operatorname{Re}\left( z \right)+\operatorname{Im}\left( z \right)=\frac{z+\bar{z}}{2}+\frac{-iz+i\bar{z}}{2}=\frac{1-i}{2}z+\frac{1+i}{2}\bar{z}$  and

$\operatorname{Re}\left( z \right)-\operatorname{Im}\left( z \right)=\frac{z+\bar{z}}{2}-\left( \frac{-iz+i\bar{z}}{2} \right)=\frac{1+i}{2}z+\frac{1-i}{2}\bar{z}$

Let $A=\sum\limits_{k=1}^{n}{\left( \operatorname{Re}\left( {{z}_{k}} \right)-\operatorname{Im}\left( {{z}_{k}} \right) \right)}\,\,\,\And \,\,\,B=\sum\limits_{k=1}^{n}{\left( \operatorname{Re}\left( {{z}_{k}} \right)+\operatorname{Im}\left( {{z}_{k}} \right) \right)}$

So $A=\sum\limits_{k=1}^{n}{\left( \left( \frac{1+i}{2} \right){{z}_{k}}+\left( \frac{1-i}{2} \right){{{\bar{z}}}_{k}} \right)=\left( \frac{1+i}{2} \right)\sum\limits_{k=1}^{n}{{{z}_{k}}+\left( \frac{1-i}{2} \right)\sum\limits_{k=1}^{n}{{{{\bar{z}}}_{k}}}}}$ and

$B=\sum\limits_{k=1}^{n}{\left( \left( \frac{1-i}{2} \right){{z}_{k}}+\left( \frac{1+i}{2} \right){{{\bar{z}}}_{k}} \right)=\left( \frac{1-i}{2} \right)\sum\limits_{k=1}^{n}{{{z}_{k}}+\left( \frac{1+i}{2} \right)\sum\limits_{k=1}^{n}{{{{\bar{z}}}_{k}}}}}$

So $\left| A \right|=\left| \left( \frac{1+i}{2} \right)\sum\limits_{k=1}^{n}{{{z}_{k}}+\left( \frac{1-i}{2} \right)\sum\limits_{k=1}^{n}{{{{\bar{z}}}_{k}}}} \right|\le \left| \frac{1+i}{2} \right|\left| \sum\limits_{k=1}^{n}{{{z}_{k}}} \right|+\left| \frac{1-i}{2} \right|\left| \sum\limits_{k=1}^{n}{{{{\bar{z}}}_{k}}} \right|$

But $\left| \frac{1+i}{2} \right|=\left| \frac{1-i}{2} \right|=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$ and $\left| {{z}_{k}} \right|=\left| {{{\bar{z}}}_{k}} \right|$ so $A\le \frac{2}{\sqrt{2}}\sum\limits_{k=1}^{n}{\left| {{z}_{k}} \right|}$

Similarly , $\left| B \right|\le \frac{2}{\sqrt{2}}\sum\limits_{k=1}^{n}{\left| {{z}_{k}} \right|}$ Thus $\left| AB \right|\le \left| A \right|\left| B \right|=2{{\left( \sum\limits_{k=1}^{n}{\left| {{z}_{k}} \right|} \right)}^{2}}$

Therefore $\frac{\left| AB \right|}{{{\left( \sum\limits_{k=1}^{n}{\left| {{z}_{k}} \right|} \right)}^{2}}}\le 2$                                                       Q.E.D

*___________________________
The idea of Solution credit to Ravi Prakash
        and

Diego Alvariz