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#: <math>\mathbf{a}\cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta = \|\mathbf{b}\|\|\mathbf{a}\|\cos\theta = \mathbf{b}\cdot\mathbf{a} </math>

# '''[[Distributif property|Distributif]] over vector addition:'''

# '''[[bilinear form|Bilinear]]''':

#: <math> \mathbf{a} \cdot (r\mathbf{b} + \mathbf{c})

:<math>\mathbf{a}\cdot \mathbf{b} = \sum{a_i \overline{b_i}} </math>

where <span style="text-decoration: overline">''b_i''</span> is the [[complex conjugate]] of ''b_i''. Then the scalar product of any vector with itself is a non-negative real number, and it is nonzero except for the zero vector. However this scalar product is thus [[sesquilinear]] rather than bilinear: it is [[conjugate linear]] and not linear in '''b''', and the scalar product is not symmetric, since

The angle between two complex vectors is then given by

:<math>\cos\theta = \frac{\operatorname{Re}(\mathbf{a}\cdot\mathbf{b})}{\|\mathbf{a}\|\,\|\mathbf{b}\|}.</math>