Is there an extension of the complex numbers in which $|z|^2 = -1$ admits a solution? Just as $x^2 = -1$ required the invention of imaginary numbers, I can think of another equation ie $|z|^2 = -1$ where $z=a+ib$. For this, don't we need to go to a higher dimension space (i.e., 3) and define a more general set?
 A: It depends on what one means, exactly:
First note that the two equations are not quite analogous: Unlike $x^2 = -1$, the equation $|z|^2 = - 1$ is not a polynomial equation over the underlying division algebra $\Bbb C$ in which the variable lives. (Indeed, $|z|^2 = z \bar{z}$, but $\bar{z}$ is not polynomial in $z$.)
Next, $| \cdot |$ is a norm, which by definition is a map $\Bbb C \to [0, \infty)$, so if we wanted to realize $(\Bbb C, |\cdot|^2)$ as the restriction of some structure $(\Bbb A, N(\cdot))$ (1) for which the restriction of $N(\cdot)$ to $\Bbb C$ coincides with $|\cdot|^2$, and (2) that admits solutions to $N(q) = -1$, we cannot insist that $N(u)$ is the square of a norm. In particular, under the extension from $\Bbb R$ to $\Bbb C$ in the question, the extension from $\Bbb C$ to $\Bbb A$ cannot be an inclusion of division algebras endowed with the appropriate norm (or quadratic form).
If we're willing to tolerate this loosening, however, we can readily find such a structure (and a little work shows that it is familiar), albeit in dimension $4$ (over $\Bbb R$): Using the Generalized Cayley-Dickson Construction, we can define an algebra $\widetilde{\Bbb H}$ on the underlying set $\Bbb C \times \Bbb C$ by
$$(p, q) \cdot (r, s) := (pr + \bar{s} q, sp + q\bar{r}),$$ and this algebra enjoys a conjugation operator $\widetilde{\Bbb H} \to \widetilde{\Bbb H}$ defined by $$\overline{(p, q)} := (\bar{p}, -q) .$$ This structure is sometimes called the split-quaternions and shares some properties with the usual algebra $\Bbb H$ of quaternions. The conjugation determines a quadratic form $\widetilde{\Bbb H} \to \Bbb R$ by $$N(u) := u \bar{u} = \bar{u} u$$ that makes $\widetilde{\Bbb H}$ a composition algebra. This requires a little explanation: For any $u$, $N(u)$ has the form $(a, 0)$ for some $a$ in $\Bbb R$, and declare interpret $N(u)$ as the real number $a$. Computing gives
$$N(p, q) = (p, q) \overline{(p, q)} = (p, q)(\bar{p}, -q) = p \bar{p} - q \bar{q} = |p|^2 - |q|^2 ,$$ so in particular $N$ is not positive definite, and hence not the square of some norm.
Now, we claim that the structure $(\widetilde{\Bbb H}, N(\cdot))$ satisfies the criteria (1) and (2):
(1) Consulting the above formulas shows that we can embed the algebra $\Bbb C$ 
via the map $z \mapsto (z, 0)$. In particular, this respects multiplication and composition, and $N(z) = |z|^2$.
(2) By the above formula, we have $N(0, 1) = |0|^2 - |1|^2 = -1$.
It turns out that $\widetilde{\Bbb H}$ is isomorphic (as an $\Bbb R$-algebra) to the algebra $M(2, \Bbb R)$ of $2 \times 2$ matrices. In this realization, the quadratic form $N$ is the determinant, $$\det : \pmatrix{a&b\\c&d} \mapsto ad - bc,$$ conjugation is the adjugate map $$\operatorname{adj} : \pmatrix{a&b\\c&d} \mapsto \pmatrix{d&-b\\-c&a},$$ and we can regard the embedding of $\Bbb C$ as the usual identification $$z = a + i b \leftrightarrow \pmatrix{a&-b\\b&a}.$$
