Is the Scalar Product Definition in my book wrong? In Rainer Kress'es book "linear integral equations" (2nd edition) on page 9 it says

Definition 1.19 Let X be a complex (or real) linear space. Then a function $(\cdot , \cdot): \rightarrow X \times X \rightarrow \mathbb{C}\; (or \mathbb{R})$ with the properties.
  (H1) $(\varphi, \varphi) \geq 0 $ (positivity)
  [...].
  for all $\varphi \in X$ is called a Scalar product.  

Now, if the mapping goes from $X\times X$ to $\mathbb{C}$, we might have to compare imaginary numbers with the > relation, which is not possible to my knowledge.
Is this a mistake in the book, or did I miss something?
Edit: pardon my formatting, I'm typing this on my phone
Clarification
Why do we map $X \times X \rightarrow \mathbb{C}$ in the first place if we implicitly assume it is real anyway. I find this confusing.
Result
My confusion came from, that I thought of a mapping to be a  Scalar product which turned out it was not. 
 A: It is correct, because the only elements of $\mathbb C$ for which the relation $\geq$ is defined are real numbers. Therefore, the statement $$z\in\mathbb C\land z \geq 0$$ is equivalent to the statement $$z\in\mathbb R \land z\geq 0.$$

We do not implicitly assume that the mapping is real. We only assume that $(\phi, \phi)$ is real for all values of $\phi$, not that $(\phi, \psi)$ is real for all values of $\phi,\psi$.
For example, the mapping $$(.,.):\mathbb C^2\to \mathbb C\\ (z,w)\mapsto z\cdot \overline w$$
is a scalar product, and $(z,z)$ is always real, however $(i,1)$ is not real.
A: Without context, it is difficult to say if there may have been a mistake. However, if done correctly, one of the other properties of $(\cdot, \cdot)$ will be the fact that $(\phi, \psi) = \overline{(\psi,\phi)}$, where the line denotes complex conjugation. In particular, you have that $(\phi,\phi) = \overline{(\phi,\phi)}$ and thus $(\phi, \phi)$ is always real.
Alternatively, for complex numbers $\alpha$, you could interpret $\alpha \geq 0$ as meaning "$\alpha$ is real, and furthermore $\alpha \geq 0$".
A: It follows implicitly that $(\phi,\phi)$ must be a real number and be $\ge 0$ at the same time. This is implicit here. Wikipedia is a bit more explicit.
A: A scalar product is, as the name implies, a product whose result is a scalar. If X is a complex linear space, then the space of scalars will be the complex numbers, so the scalar product will map to the complex numbers. So the codomain of the scalar product will be the set of complex numbers, even though the range will be positive real numbers for a vector with itself.
