Hyperplane in a complex vector space My friend, who studies Physics, asked me about the meaning of "functional" so I gave the definition and some examples. To motivate its importance, I explained how a functional can be use to define a hyperplane without referring to a specific base of the space (A subset $H$ is a hyperplane iff there exists a non-trivial linear functional $x'$ and a scalar $c$ such that $x'(x) = c$ for all $x \in H$ ) and that it effectively divides the space into 3 parts e.g. $x'(x) < c$, $x'(x) = c$, and $x'(x) > c$. I immediately notice that the argument works in real vector spaces but not the complex ones since complex numbers are not linearly ordered, thus the intuitive picture that hyperplanes "divide space" in the aforementioned sense seems to fail here.
So, is there an intuitive way to visualize a complex hyperplane? For concreteness, you can assume that the space is a finite dimensional Hilbert space. Note that I am an undergraduate so I'd really appreciate some not too advanced answers (stuff like Hopf fibration would be considered too advanced for me, for example).
 A: Since this has gotten bumped, and may be useful to posterity:
Literally, a complex hyperplane in a finite-dimensional complex vector space is (i) a real subspace of real codimension two that (ii) is closed under complex scalar multiplication. How are we to think about this geometrically?

(i) The picture of a real hyperplane $H$ dividing the ambient space $V$ into two half-spaces can be understood by picking a complementary real one-dimensional complement $N$. (In an inner product space we'd often take $N = H^{\perp}$ to be the orthogonal complement, but that's not necessary.) Since $V = H \oplus N$, we can partition $V$ into cosets $H + \{n\}$, namely affine translates of $H$ by elements $n$, which are parametrized by elements of $N$ in an obvious sense. Intuitively, $H$ separates $V$ because $\{0\}$ separates the space of cosets, namely the real line $N \simeq V/H$.
The analogous picture makes sense for complex hyperplanes. One crucial difference is, the normal space $N$ is a complex one-dimensional subspace of $V$, so the local structure of the quotient $V/H$ is that of a point in a complex line (or a real plane). Particularly, there exist closed loops in $V$ that do not cross $H$, but that "link" or "wind around" $H$ in the same way the unit circle winds around $0$ or links a coordinate axis in real Cartesian three-space.
In more detail, identify $(z, w) = (x + iy, u + iv)$ in $\mathbf{C}^{2}$ with $(x, y, u, v)$ in $\mathbf{R}^{4}$. The complex hyperplane $H = \{z = 0\}$ is identified with the real two-plane $\{x = y = 0\}$. We may pick as normal space the complex line $N = \{w = 0\}$, which contains the circle
$$
\{(z, w) : |z| = 1,\ w = 0\} = \{(e^{i\theta}, 0) : \text{$\theta$ real}\}.
$$
In the real three-dimensional hyperspace $\{v = 0\}$, a.k.a., Cartesian three-space with coordinates $(x, y, u)$, this circle is our old friend The Unit Circle, the only part of $H$ we can see is its intersection with $\{v = 0\}$ (namely the $u$-axis), and the circle links $H$ in the sense of the preceding paragraph.
(ii) Being closed under complex scalar multiplication is deceptively simple. The simplest non-trivial situation is $V = \mathbf{C}^{2}$, a real four-dimensional space. Each non-zero vector $v$ in $V$ determines a unique complex line $H$, spanned as a real two-plane by the set $\{v, iv\}$.
Now, "most" vectors $w$ in $V$ do not lie in $H$. The real two-plane spanned by the set $\{v, w\}$ is therefore "usually" not equal to $H$,[*] and is therefore not a complex hyperplane. The point is, complex hypersurfaces really are Very Special among real subspaces of real codimension two.
[*] In fancy terms, there is a real two-sphere's worth of complex hyperspaces, the complex projective line, but a product-of-two-spheres' worth of real two-planes, the unoriented Grassmannian.
