Is a complex space more "advanced" than a "generic" real space? For instance, does taking the square root of a complex number and its complex conjugate create a metric that "automatically" makes it an inner product space?
Is a complex space more complete than a real space? If not, what must be done to make it complete, so that it IS a Hilbert space?
And does the modulus represent a norm that makes a complete complex space a Banach space? 
 A: We can define an inner product on $\mathbb{C}$ by the rule $\langle z,w \rangle = z\overline{w}$ for all $z,w\in\mathbb{C}$. The norm on $\mathbb{C}$ induced by this inner product is then the map $z\to \left|z\right|$ where $\left|z\right|$ denotes the modulus of the complex number $z$. Finally, $\mathbb{C}$ is complete under the norm induced by this inner product and is therefore a Hilbert space.
Additional Details:
A vector space $V$ over a field $F$ ($F=\mathbb{R}$ or $F=\mathbb{C}$) is an inner-product space if there is a map $V\times V\to F$ (for notational convenience we denote the image of $(z,w)$ under this map by $\langle z,w \rangle$) that satisfies the following axioms:
(1) $\langle v,v \rangle \geq 0$ for all $v\in V$ and $\langle v,v \rangle =0$ if and only if $v=0$.
(2) If $u,v,w\in V$, then $\langle u+v,w \rangle  =\langle u, w\rangle + \langle v,w \rangle$.
(3) If $a\in F$ and if $u,v\in V$, then $\langle au,v\rangle = a\langle u,v\rangle$.
(4) If $u,v\in V$, then $\langle u,v\rangle =\overline{\langle v,u\rangle}$.
Exercise 1: Prove that the inner product on $\mathbb{C}$ given by the rule described at the very beginning of this answer is indeed an inner product, that is, it satisfies axioms (1)-(4) above.
Note that axiom 4 can be removed if $F=\mathbb{R}$. If $V$ is an inner product space, then the norm induced by the inner product on $V$ is the map $v\to \sqrt{\langle v,v\rangle}$. (The image of $v\in V$ under this map is denoted by $\left\|v\right\|$ for notational convenience.)
Exercise 2: Prove that the norm induced by the inner product on $\mathbb{C}$ given by the rule described at the very beginning of this answer is indeed the map $z\to \left|z\right|$ where $\left|z\right|$ denote the modulus of the complex number $z$.
If $V$ is an inner product space, then we can define a metric $d:V\times V\to [0,\infty)$ by the rule $d(u,v)=\left\|u-v\right\|$ for all $u,v\in V$. 
Exercise 3: Prove that $d$ is indeed a metric on $V$.
Finally, a Hilbert space is an inner-product space $V$ that is complete under the metric induced by its norm.
Exercise 4: Prove that $\mathbb{C}$ is a Hilbert space under the inner product described at the very beginning of this answer.
I hope this helps!
A: Here's a compendium of answers by others, from the comments:
Connections between metrics, norms and scalar products (for understanding e.g. Banach and Hilbert spaces)
Norms Induced by Inner Products and the Parallelogram Law
You should be aware of the following constructions for a vector space:
inner product ⇒ norm ⇒ metric ⇒ topology
That is, given an inner product, there is an induced norm, but not necessarily vice versa, and so on. – 
