- Extending the notion of points and lines

- Properties of imaginary and real elements

- A real line and two conjugate imaginary points.

- Two conjugate imaginary lines determine a real point.

- A real line contains an infinity number of imaginary points

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

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

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

- Real points on an imaginary line.

- Real lines through an imaginary point.

- Theorem 1

- Theorem 2

- A real line and two conjugate imaginary points.
- Real and imaginary curves - definitions

- Isotropic or cyclic points and lines

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||.

- 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. - 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.

- real regular points; Example: (2,4,5)
- real ideal points; Example: (2,4,0)
- imaginary regular points; Example: (2+i,4,5i)
- imaginary ideal points; Example: (2+i,4,0)

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

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.

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.

- real regular lines; Example: 2x + 4y + 7z = 0
- the real ideal line z = 0
- imaginary regular lines; Example: (2+i)x + 4y - 5iz = 0

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

Two different lines have exactly one common point.

Two different points determine exactly one line.

All formulas about equations of lines, points, ... remain unchanged.

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.

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.

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.

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.

The proof is left as an exercise

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=0Similarly, 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.

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

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

Example:

xare equations of imaginary curves.^{2}+ y^{2}= 0 and x^{2}+ y^{2}+ 9 = 0

These points are ideal and conjugate imaginary.

Each line that contains such point is called an isotropic line.

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 relative to the old axes.

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

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

[1 0 xo] [ 1 ] [ 1 ] [0 1 yo].[ i ] = [ i ] [0 0 1] [ 0 ] [ 0 ]relative to the new axes, take the point J(1,-i,0). The coordinates of this point relative 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.

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]relative to the new axes, take the point I(1,-i,0). The coordinates of this point relative 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)-i sin(t))(cos(t)+i sin(t)),(sin(t)+i cos(t))(cos(t)+i sin(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.

The tutorial address is http://home.scarlet.be/math/

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