How complex numbers can be compared in Cauchy-Schwarz inequality in an inner product space over the complex field

Question

The Cauchy-Schwarz inequality is given by $|\langle x,y \rangle| \leqslant \|x\| \|y\|,\, x,y\in V$ . My question is that over the complex fields the RHS of the inequality will be a complex number. But we know that complex numbers cannot be compared. Then how is it possible to compare them in this inequality?

Answer 1

In $\Bbb C^n$, the RHS is a non-negative real number, since$$\|(z_1,z_2,\ldots,z_n)\rVert=\sqrt{|z_1|^2+|z_2|^2+\cdots+|z_n|^2}.$$