Imaginary points and lines

Extending the notion of points and lines

Points

In the theory about homogeneous coordinates, we have seen that with each set ||a,b,c||, there is exactly one point.
The values a, b and c were real numbers.

Now, we extend the definition of ||a,b,c||.
S is the set of all ordered triples (a,b,c) ; with a,b,c in the set C of complex numbers.
Now we remove the element (0,0,0). So = S \ {(0,0,0)}.
In So, we say that (a',b',c') is a multiple of (a,b,c) if and only if there is a complex number s such that (a',b',c') = s (a,b,c).
In So, we denote the set of all these multiples of (a,b,c) as ||a,b,c||.

Now we'll associate just one point with each set ||a,b,c||.

  1. 1.      If ||a,b,c|| contains at least one triple real numbers (r,r',r").
    (r,r',r") are the coordinates of one real point P.
    We associate ||a,b,c|| with that point P.
    Each element of that set is a triple homogeneous coordinates of point P.
    Example: P(i,i,5i) is a real point.
  2. 2.      If ||a,b,c|| contains no triple real numbers (r,r',r").
    We associate ||a,b,c|| with a imaginary point. We say that each element of the set ||a,b,c|| is a triple homogeneous coordinates of that point. At least one coordinate is not real.
    For instance, we can talk about the point P(1+i,5,-i).

Now we have four sorts of points.

Conjugate imaginary points

 
        P(a+ia', b+ib', c+ic') and point Q are conjugate imaginary points
<=>
        (a-ia', b-ib', c-ic') are coordinates of Q

Criterion for real points

If a, b and c are real, then obviously P(a,b,c) is real.
If a, b and c are real, then obviously P(ia,ib,ic) is real.

Now we consider all other cases.

 
        P(a+ia',b+ib',c+ic') is a real point
=>
        ||a+ia',b+ib',c+ic'|| contains a real triple (d,e,f)
=>
        There is a complex number k + il such that
                a+ia' = (k+il)d
                b+ib' = (k+il)e
                c+ic' = (k+il)f
=>
        a = kd          a' = ld
        b = ke   and    b' = le
        c = kf          c' = lf
=>
        (a,b,c) is directly proportional with (d,e,f) and
        (a',b',c') is directly proportional with (d,e,f)
=>
        (a,b,c) is directly proportional with (a',b',c')
=>
        There is a real value t such that
         a' = ta and b' = tb and c' = tc
=>
        point P has coordinates (a+ita,b+itb,c+itc)
=>
        point P has coordinates  (a(1+it),b(1+it),c(1+it))

Conversely:
If point P has coordinates (a(1+it),b(1+it),c(1+it)), then P has coordinates (a,b,c) and therefore P is real.

Lines

From the theory about homogeneous coordinates, we know:
With each line a with equation u x + v y + w z = 0 corresponds exactly one set ||u,v,w||.
Each element of that set is called 'line coordinates' of line a.
So, we can write : line a(u, v, w)
(u,v,w) are homogeneous coordinates of the line.

In exactly the same way as above we define real lines, imaginary lines and conjugate imaginary lines.

Now we have three sorts of lines.

Criterion for real lines

Same criterion as above.

Definition

For all points and all lines, we have: 

 
        Point P(a,b,c) is on line l(u,v,w)
<=>
         u a + v b + w c = 0

Corollaries

It is easy to prove that :
Two different lines have exactly one common point.
Two different points determine exactly one line.
All formulas about equations of lines, points, ... remain unchanged.

Properties of imaginary and real elements

A real line and two conjugate imaginary points.

Two conjugate imaginary points determine a real line.
Take P(a+ia',b+ib',c+ic') and Q(a-ia',b-ib',c-ic').
The line PQ has equation 

 
        | x             y               z       |
        | a+ia'         b+ib'           c+ic'   | = 0
        | a-ia'         b-ib'           c-ic'   |
 
<=>     (row 2 + row 3)
 
        | x             y               z       |
        | 2a            2b              2c      | = 0
        | a-ia'         b-ib'           c-ic'   |
 
<=>     (row 3 - (1/2) row 2) and dividing by (-i)
 
        | x     y       z  |
        | a     b       c  | = 0
        | a'    b'      c' |

and this is a real line.

Two conjugate imaginary lines determine a real point.

The proof is left as an exercise

A real line contains an infinity number of imaginary points

Take the real and different points P(a,b,c) en Q(a',b',c') on a real line.
The point R(a+ita',b+itb',c+itc'), with t as a real parameter, lies on PQ and is an imaginary point because (a,b,c) and (a',b',c') are not directly proportional.
If t varies, an infinity number of imaginary points arise.

A real point is on an infinity number of imaginary lines.

The proof is left as an exercise

If a real line contains an imaginary point, that line contains the conjugate imaginary point.

Say that the real line a contains the imaginary point P.
Suppose that the conjugate imaginary point P' is not on line a, then the real line PP' and the real line a intersect in a real point. This is impossible since P is an imaginary point.

If an imaginary line contains a real point P, then the conjugate imaginary line contains point P.

The proof is left as an exercise

Real points on an imaginary line.

There is exactly one real point on an imaginary line.
The imaginary line and the conjugate imaginary line intersect at a real point. Thus, there is a real point on an imaginary line.
If there are two real points on a line, it is a real line.

Real lines through an imaginary point.

There is exactly one real line through an imaginary point.
The proof is left as an exercise

Theorem 1

If
R is a real point
P and Q are conjugate imaginary points
P, Q and R are not collinear
Then
The lines RP and RQ are conjugate imaginary lines

Proof:
Take P(a+ia',b+ib',c+ic') and Q(a-ia',b-ib',c-ic') and R(d,e,f)
The equation of RP is 

 
        | x             y               z     |
        | d             e               f     | = 0
        | a-ia'         b-ib'           c-ic' |
<=>
        | x     y       z  |     | x    y   z  |
        | d     e       f  | + i | d    e   f  | = 0
        | a     b       c  |     | a'   b'  c' |
<=>
        (ux+vy+wz)+i(u'x+v'y+w'z)=0
<=>
        (u+iu')x+(v+iv')y+(w+iw')z=0

Similarly, the equation of RQ is (u-iu')x+(v-iv')y+(w-iw')z=0
Both lines are not real lines because if RP is real, then Q is on RP and P, Q and R are collinear. This gives a contradiction.

Theorem 2

If
r is a real line
p and q are conjugate imaginary lines
P, q and r are not concurrent
Then
The intersection points of r and p, and of r and q, are conjugate imaginary points.

The proof is left as an exercise

Real and imaginary curves - definitions

We say a curve is real if and only if it contains an infinity number of real points.

We say a curve is imaginary if and only if it contains a finite number of real points.
Example: 

 
        x2  + y2  = 0  and  x2  + y2  + 9 = 0

are equations of imaginary curves.

Isotropic or cyclic points and lines

Definitions

With respect to orthonormal axes we define the points I(1,i,0) and J(1,-i,0) as cyclic or isotropic points.
These points are ideal and conjugate imaginary.
Each line that contains such point is called an isotropic line.

Translation and isotropic points

Choose a translation from an old system of axes to a new system.
From the theory about coordinate transformations, we know that the transformations formulas are 

 
        [ x ]     [ x' ]                [1   0   xo]
        [ y ] = M.[ y' ]  with      M = [0   1   yo]
        [ z ]     [ z' ]                [0   0    1]

(x,y,z) are the coordinates of a point with respect to the old axes.
(x',y',z') are the coordinates of a point with respect to the new axes.

With respect to the new axes, take the point I(1,i,0). The coordinates of this point with respect to the old axes are 

 
        [1   0   xo] [ 1 ]   [ 1 ]
        [0   1   yo].[ i ] = [ i ]
        [0   0    1] [ 0 ]   [ 0 ]

With respect to the new axes, take the point J(1,-i,0). The coordinates of this point with respect to the old axes are 

 
        [1   0   xo] [ 1 ]   [ 1 ]
        [0   1   yo].[ -i] = [ -i]
        [0   0    1] [ 0 ]   [ 0 ]

Conclusion: The coordinates of the cyclic points are invariant with respect to a translation.

Rotation and isotropic points

Choose a rotation from an old orthonormal system of axes to a new orthonormal system.
From the theory about coordinate transformations, we know that the transformations formulas are 

 
        [ x ]     [ x' ]                [cos(t)   -sin(t)   0]
        [ y ] = M.[ y' ]  with      M = [sin(t)    cos(t)   0]
        [ z ]     [ z' ]                [  0         0      1]

With respect to the new axes, take the point I(1,-i,0). The coordinates of this point with respect to the old axes are 

 
        [cos(t)   -sin(t)   0] [ 1 ]  [cos(t) - i sin(t) ]
        [sin(t)    cos(t)   0].[ i ] =[sin(t) + i cos(t) ]
        [  0         0      1] [ 0 ]  [       0          ]

We multiply this coordinates with (cos(t) + i sin(t)) 

 
 ((cos(t)-isin(t))(cos(t)+isin(t)),(sin(t)+icos(t))(cos(t)+isin(t)), 0)
<=>
        ...
<=>
        ( 1 , i , 0 )

You'll find a similar result for J(1,-i,0).
Conclusion: The coordinates of the cyclic points are invariant with respect to a rotation.