Adding a root of $z\bar z=-1$ to $\mathbb C$ This half-serious question is inspired by the answer to my previous one, Want something like Cayley formula for unitary matrices
The equation $z^2=-1$ does not have solutions in $\mathbb R$; adding a solution produces $\mathbb C$.
The equation $z\bar z=-1$ does not have solutions in $\mathbb C$; adding a solution produces what?
Update - having learned more thanks to the comments and the answer, I've now posted a question on MO with hopefully more serious and interesting content in it: https://mathoverflow.net/q/248241/41291

It seems, the answer depends on the definition of conjugation. The answer by @quid seems to be isomorphic to bicomplex numbers or tessarines, except for the different definition of conjugation.
If we consider the ring of tessarines, with conjugation being defined as changing the sign of all occurrences of $i$ to $-i$, there is no solution. This definition of conjugation is equivalent as transposition of tessarines represented as $4\times4$ real matrices or conjugate-transpose of tessarines represented as $2\times2$ complex matrices.
In other words, we add $j$ such that $j^2=1$, $\overline{j}=j$, $(ij)^2=-1$, $\overline{ij}=-ij$
Code in Mathematica, proving there is no solution:
Solve[Dot[( {
     {w0, -w1, w2, -w3},
     {w1, w0, w3, w2},
     {w2, -w3, w0, -w1},
     {w3, w2, w1, w0}
    } ), Transpose[( {
      {w0, -w1, w2, -w3},
      {w1, w0, w3, w2},
      {w2, -w3, w0, -w1},
      {w3, w2, w1, w0}
     } )]] == ( {
    {-1, 0, 0, 0},
    {0, -1, 0, 0},
    {0, 0, -1, 0},
    {0, 0, 0, -1}
   } ), {w0, w1, w2, w3}, 
 Assumptions -> 
  w0 \[Element] Reals && w1 \[Element] Reals && w2 \[Element] Reals &&
    w3 \[Element] Reals]
 
Out={ }

