Why is dot product for complex numbers as such To my knowlege the inner product is defined as
$$ \langle a, b \rangle = \sum_{i=1}^{n} a_i\bar{b_i} \,.$$
While the complex conjugate doesn't have an effect on real numbers, it does effect the dot product for scalar numbers. What is the point of the conjugate... It doesn't quite make sense? Would someone be willing to explain why its the complex conjugate instead of just multiplying the entries?
Thanks!
 A: One way to motivate the definition of the complex scalar product is to think about how the scalar product relates to norms. In the real case we have that the euclidean norm is given by
$$ \| x \|_2 = \left( \sum_{i=1}^n x_i^2 \right)^{1/2}
   \quad\text{for}\quad x \in \mathbb{R}^n
   \,. $$
In the complex case, we can replace the square of the entry of the vector, by the square of the absolute value of the complex number to obtain a reasonable definition for the norm, i.e.,
$$ \| z \|_2 = \left( \sum_{i=1}^n | z_i |^2 \right)^{1/2}
   \quad\text{for}\quad z \in \mathbb{C}^n
   \,. $$
Now, we want that for our complex scalar product $\langle z, z \rangle^{1/2} = \| z \|$ holds. Note, $| z_i |^2 = z_i \overline{z_i}$, and thus we can see that by defining
$$ \langle z, w \rangle = \sum_{i = 1}^n \overline{w_i} z_i $$
the desired equality holds. We would, of course, still need to check that this definition fulfills all the scalar product axioms.
Finally, you can use the polarization identity to show that this scalar product is the only one that induces the norm as defined above.
