# Complex Numbers on Elliptic Curves & Usage in Tate Pairing

I'm working with understanding the internals of the Tate Pairing. I was going through an example of the curve $E: y^2 = x^3 + 3x$ over $\mathbb{F_{11}}$. The author is showing the computation of $e(P,Q)$, where $P=(1,9)$ and $Q = \phi (P) = (10,9i)$ using Miller's Algorithm. Through this algorithm, the author chose a point $Q' = (6,6)$ and then computes $S = Q + Q' = (8+7i, 10+6i)$. This is where I'm having an issue, I can't seem to determine how the complex addition of points occurs in elliptic curves, and I'm looking for an explanation in the usage for the 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 &\equiv (a_0 x + b_0) \cdot (a_1 x + b_1) \\ &\equiv a_0 a_1 x^2 + a_0 b_1 x + b_0 a_1 x + b_0 b_1 \\ &\equiv a_0 a_1 \cdot (-1) + a_0 b_1 x + b_0 a_1 x + b_0 b_1 \\ &\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 $$\mathbb F_{11^2} \cong \mathbb F_{11}[i]/(i^2 + 1)$$.