# Complex number being grater than $0$ in the definition of inner product [duplicate]

I was reading the definition of inner product here and they state the following:

Let $$V$$ be a vector space over a field $$\mathbb F$$ ( $$\mathbb R$$ or $$\mathbb C$$). A map $$\left< \cdot , \cdot \right>: V \times V \to \mathbb F$$ is called an inner product if, for all $$x,y,z \in V$$ and $$\alpha \in \mathbb F$$:

1. $$\left< \alpha x , y \right> = \alpha \left< x , y \right>$$ and $$\left< x + z , y \right> = \left< x , y \right> + \left< z , y \right>$$
2. $$\left< x , y \right> = \overline{\left< y , x \right>}$$
3. $$\left< x , x \right> > 0$$, if $$x \neq 0$$.

My question is about property 3. Let $$V$$ be a vector space over $$\mathbb C$$, then $$\left< \cdot , \cdot \right>: V \times V \to \mathbb C$$, this means that for all $$x \in V$$, $$\left< x , x \right>$$ is a complex number. So what do they mean by $$\left< x , x \right> > 0$$? Do they mean that the inner product of $$x$$ with itself is allawys a real number, so if $$\left< \cdot , \cdot \right>$$ is an inner product then $$\Im(\left< x , x \right>) = 0$$? If not, what does it mean for a complex number to be greater than $$0$$ in this context?

Because of the second property, $$\langle x,x \rangle = \overline{\langle x,x \rangle}$$, so $$\Im(\langle x,x \rangle)=-\Im(\langle x,x \rangle)=0$$, i.e. $$\langle x,x \rangle$$ is real.
$$\left< x , y \right> = \overline{\left< y , x \right>}$$ applied for $$x= y$$, you get
$$\left< x , x \right> = \overline{\left< x , x \right>}.$$
Therefore $$\left< x , x \right>$$ is a real number.