#### Conics

As a warm up, we start with a standard example y^{2} = x. When x,y are treated as
real variables, this is simply a parabola opening sideways. Things get more
interesting when the variables are allowed to take complex values. Since the graph
would live in 2 complex or 4 real dimensions, it is impossible visualize completely.
Nevertheless, we can get a sense of it projecting to 3 real dimensions, by setting x_{1} and x_{2} to
be real and imaginary parts of x, and x_{3} to be the real part of y. ( We will use
x_{i} as coordinates of R^{3}.)
In polar
coordinates, this gives

We can now plot the graph, using colour to encode the imaginary part of
y.
Here θ runs from 0 to 4π.
Note that even though the surface appears to cross itself. We can see that it doesn't since the colours of sheets are different
where they meet.
In general, a * conic * is given by a quadratic equation f(x,y) = 0.
The standard classification that one learns in highschool becomes
simpler over the complex numbers, because now there is no difference
between an ellipse and a hyperbola. To clarify this, we say that two conics
are equivalent if we can transform one to the other by an
affine transformation

where we require the abde matrix to be invertible.
Then up to equivalence, there are only 4 possibilities
- A union of two lines.
- A double line.
- The parabola
y
^{2} = x.
- The circle
x
^{2} + y^{2} = 1

We want to give a complex plot of the circle.
We start with the usual parameterization

and
then substitute θ = s + it. From standard identities, we obtain
Plotting real and imaginary parts of x and the real part of y yields
The above parameterization of the circle uses transcendental functions. There is
another, perhaps less well known, parameterization using only rational functions.
Naturally this is referred to as a *rational parameterization*. The idea is choose a point
P on the circle, and parameterize the lines through it by their slope t. These should
cut the circle in another point, which can then be expressed in terms of t. Picking
P = (0,1), we see that t = (y - 1)∕x. Solving for y and then substituting into the
equation for the circle gives

Supposing x is nonzero as is typical, we get
We didn’t use these formulas
for the plotting, because they are less convenient (we would need
t = ∞ to complete the circle). However, it has other uses. Suppose that we wanted
find all right angle triangles with integer sides. By
Pythagoreas’ theorem, this equivalent to finding positive
integer solutions for a^{2} + b^{2} = c^{2}. Via the substitution x = a∕c, y = b∕c, this is
amounts to finding points on the circle with positive rational coordinates. Here’s a
very simple procedure: Choose a rational number -1 < t < 0, and substitute into the
above formulas.

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