# Norm of the sum of complex vectors. Say that we have n complex-valued vectors z_i and we want to evaluate: norm(sum_(i=1)^n z_i)_2^2

Say that we have n complex-valued vectors ${\mathbf{z}}_{i}$ and we want to evaluate:
${‖\sum _{i=1}^{n}{\mathbf{z}}_{i}‖}_{2}^{2}$
Now, I know that for real vectors ${\mathbf{x}}_{i}$ it holds:
${‖\sum _{i=1}^{n}{\mathbf{x}}_{i}‖}_{2}^{2}=\sum _{i=1}^{n}{‖{\mathbf{x}}_{i}‖}_{2}^{2}+\sum _{i\ne j}{\mathbf{x}}_{i}\cdot {\mathbf{x}}_{j}$
But when dealing with complex vectors, the last term with the inner product will be a complex number in general so I am not sure that this formula generalizes for this case. What am I missing?
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espovilham7
For two complex numbers
$‖{z}_{1}+{z}_{2}{‖}^{2}=\left({z}_{1}+{z}_{2}{\right)}^{\ast }\left({z}_{1}+{z}_{2}\right)=‖{z}_{1}{‖}^{2}+‖{z}_{2}{‖}^{2}+\left({z}_{1}^{\ast }{z}_{2}+{z}_{1}{z}_{2}^{\ast }\right)=\phantom{\rule{0ex}{0ex}}‖{z}_{1}{‖}^{2}+‖{z}_{2}{‖}^{2}+2Re\left({z}_{1}^{\ast }{z}_{2}\right)$
easily extended to n.
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Zack Chase
Generalizing first answer to n terms, since $‖z{‖}_{2}^{2}={z}^{\ast }\cdot z$ we have
${‖\sum _{i}{z}_{i}‖}_{2}^{2}=\sum _{i}{z}_{i}^{\ast }\cdot \sum _{j}{z}_{j}=\sum _{ij}{z}_{i}^{\ast }\cdot {z}_{j}=\sum _{i}{z}_{i}^{\ast }\cdot {z}_{i}+\sum _{i\ne j}{z}_{i}^{\ast }\cdot {z}_{j}.$