Complex Numbers on Elliptic Curves & Usage in Tate Pairing
Fix a field $k$.  Consider the polynomial $x^2 + 1$ in $k[x]$.  If $x^2 + 1$ is irreducible over $k$, then obviously the quotient $k[x]/(x^2 + 1)$ is also a field, of $\#k^2$ elements if $k$ is finite.  Note that $x^2 + 1 \equiv 0 \pmod{x^2 + 1}$, so $x^2 \equiv -1 \pmod{x^2 + 1}$; hence if we take any linear polynomials $f_0 = a_0 x + b_0$ and $f_1 = a_1 x + b_1$, we can work out the product modulo $x^2 + 1$ by the standard rules of arithmetic, substituting $-1$ wherever we see $x^2$:

\begin{align}
  f_0 \cdot f_1
  &amp;\equiv (a_0 x + b_0) \cdot (a_1 x + b_1) \\
  &amp;\equiv a_0 a_1 x^2 + a_0 b_1 x + b_0 a_1 x + b_0 b_1 \\
  &amp;\equiv a_0 a_1 \cdot (-1) + a_0 b_1 x + b_0 a_1 x + b_0 b_1 \\
  &amp;\equiv (a_0 b_1 + b_0 a_1) x + b_0 b_1 - a_0 a_1
  \pmod{x^2 + 1},
\end{align}

which, as it happens, is exactly the same arithmetic as for the familiar complex numbers, because $\mathbb C \cong \mathbb R[x]/(x^2 + 1)$ can be constructed exactly this way.  Drawing on the analogy, we name the indeterminate variable $i$ rather than $x$ for the polynomials in $\mathbb F_{11}[i]$ so that we can write $i^2 = -1$ in the field $\mathbb F_{11^2} \cong \mathbb F_{11}[i]/(i^2 + 1)$.