Transcript geometrical transformations

Geometry with Complex Numbers
Mihai Caragiu
Ohio Northern University
Abstract: In the last three years, Ohio Northern University hosted a Summer
Honors Institute for gifted high school students. The week-long "Geometry
with Complex Numbers" course was offered in 2006 and 2007. The students
were not assumed to have prior knowledge of complex numbers. In this talk I
would like to share the experience we had with introducing geometrical
transformations (such as rotations, reflections and projections) with complex
numbers to talented high-school students, and we will explore ways in which
they can be used to quickly derive elegant geometrical results including (but
not limited to) the Simson's Line and the Nine Point Circle
“The course blends algebra and geometry together in order to help students
understand the interconnections between the two subjects. Students will use
experimental activities, projects and mathematical software systems to demonstrate
how geometric shapes and concepts can be realized in the complex plane.”
ACKNOWLEDGEMENTS
Dr. Donald Hunt
Dr. Harold Putt
Dr. Rich Daquila
ONU undergraduates which helped with the Summer
Camp activities, both mathematical and recreational.
CAMP ACTIVITIES (June 10-15, 2007) – OUTLINE
 Trigonometry and Geometry - basics
 Introduction to Complex Variables
 Geometrical Transformations with Complex Numbers
 Group Projects: The Nine Point Circle and The Simson Line
First we argue for the necessity of extending the set
of real numbers to create
a domain that contains solutions of quadratic equations as simple as x 2 +1=0
IMAGINARY UNIT
i  1
2
Getting used with "imaginary numbers"
is hard for non-mathematicians!...
"Not only the practical man, but also men of letters
and philosophers have expressed bewilderment at
the devotion of mathematicians to mysterious entities
which by their very name are confessed to be imaginary."
(A. N. Whitehead, An Introduction to Mathematics,
Oxford University Press, 1958, Ch.7)
SUMMARY
 COMPLEX NUMBERS AND TRANSFORMATIONS
 ALTITUDES AND ORTHOCENTERS
 THE NINE-POINT CIRCLE
 INSCRIPTIBLE QUADILATERALS
 THE SIMSON'S LINE
 A GENERALIZATION OF THE
THEOREM ABOUT THE SIMSON'S LINE
 SIMSON LINES AND EULER CIRCLES
 ON A PUTNAM PROBLEM ON PLANE ROTATIONS
FIRST CONTACT: RECTANGULAR FORM
C =  x  iy | x, y  R
 x1  iy1    x2  iy2   x1  x2  i  y1  y2 
 x1  iy1    x2  iy2    x1 x2  y1 y2   i  x1 y2  x2 y1 
For z  x  iy define
Re( z )  x,Im( z )  y, z  x  iy, z  x  y  z  z
2
2
so that...
zz
z  z z1 z1 z2
Re( z ) 
, Im( z ) 
,

2
2
2i
z2 z2
REACTANGULAR REPRESENTATION:
GEOMETRICAL CONNECTIONS
z  x  iy may be seen as
The point  x, y  in the plane
The plane vector x, y
z2
z1

Thus the transformation z
z2  z1
w given by w  z  k
is a TRANSLATION with the vector corresponding
to the complex number k .
w  z  reflection about the x  axis.
w   z reflection about the origin.
w   z reflection about the y  axis.
RECTANGULAR REPRESENTATION  ADDITION FRIENDLY  in the sense
that the addition of vectors has an obvious geometrical meaning (addition of vectors).
However, there is a nice geometrical meaning involving
multiplication and the complex conjugate function for
complex numbers in rectangular form:
P1  z1  
  z2 z1   x2  iy2    x1  iy1    x1 x2  y1 y2   i  x1 y2  x2 y1 
P2  z2  
 
2 Area  OP1P2 
OP1 OP2
CONSEQUENCES: Let A  a  ,B  b  ,..., P1  z1  , P2  z2  ,...


* The general eq. of a line perpendicular to AB is Re z  b  a   constant.


* The general eq. of a line parallel to AB is Im z  b  a   constant.
* The segments PP
1 2 and P3 P4 are perpedicular  Re   z 4  z3   z2  z1   =0.
z 4  z3
* PP
and
P
P
are
perpedicular

purely imaginary.
1 2
3 4
z2  z1
z 4  z3
* PP
and
P
P
are
parallel

purely real.
1 2
3 4
z2  z1
POLAR AND EXPONENTIAL FORMS
z  r  cos  i sin    rei  r  0,   R 2 Z 
"MULTIPLICATIVE  FRIENDLY":
z1 z2  r1r2  cos 1   2   i sin 1   2    r1r2 e  1
i   2 
.
SPECIAL CASE
i
CONSIDER COMPLEX NUMBERS u WITH u  1, u  cos   i sin   e .
u  ei 
i   
i
i

w

uz

e
re

re
i 
z  re 
THE TRANSFORMATION
w  uz
IS A ROTATION WITH ANGLE  ABOUT THE ORIGIN
"POMPEIU's PROBLEM"
With the distances PA , PB , PC from an arbitrary point P to the vertices
of an equilateral triangle ABC one can build a triangle.
1
3
 
 
Let  =  i
 cos    i sin   .
2
2
3
3
 3  1,  2    1. Consider the equilateral
triangle with vertices A  0  ,B 1 , C    and
an arbitrary point P  z  .
If we rotate P  z  with
with vertices  z , z , 

about the origin, we get P   z  . Then the sides of the triangle PPC
3
are identical with the distances z , z  1 and z   from z
to the vertices 0,1, of the equilateral triangle we considered initially.
 z  z   1 z   2 z   z  z
2
and
 z    z 1   z 1
SIMILARITY - a problem of rotations and homotheties...
Let z1, z2 , z3, w1, w2 , w3 be complex numbers. Then the triangles z1z2z3 and w1w2w3
are similar if the way of obtaining the complex segment z1z3 from z1z2 is the same
as the way of obtaining the complex segment w1w3 from w1w2.
For example, if z1z3 can be obtained from z1z2 by a rotation with an angle of  around
5
z1 followed by a dilation by a factor of 2 with respect to z1, then w1w3 can be obtained
from w1w2 by a rotation with an angle of  around w1 followed by a dilation by a
5
factor of 2 with respect to w1.
If z1z2z3 ~ w1w2w3 and if
z3  z1
w w
 a, then 3 1  a
z2  z1
w2  w1
z1z2z3 ~ w1w2w3 if and only if
z3  z1 w3  w1

z2  z1 w2  w1
GENERAL ROTATIONS IN THE PLANE
The transformation formula for a counterclockwise rotation by an
angle  about a point z0 is w  z0  ei  z  z0  .


w  ei z  1  ei z0
GENERAL FORM OF DIRECT MOTIONS
(orientation-preserving isometries)
a translation, if   2 Z , or otherwise

w  ei z  c 
c
is a rotation of angle  around

1  ei



 R1  2  z3  , if 1   2  2 Z
R
z
R
z

 1  1   2  2  


Tk (translation) if 1   2  2 Z

COMPOSITIONS: 
R   z  Tk  R    z   ; Tk R  z   R   z  

Tk1 Tk2  Tk2 Tk1  Tk1  k2



REFLECTIONS: orientation-reversing isometries
Let z , z C , z  z . Let z C. We need to find the
1 2
1 2
reflection w of z about the line through z and z .
1
2
zz
1 . Then z  z    z  z  and w  z    z  z .
Let  =
z z
1
1
1
1
 2
 2
2 1
z z 
z z 



1
2 1  z  z 
Thus w  z    z  z  
z z 
1
1 z  z  2 1 z  z 
1
 2
2 1
2 1

z z 
w  z  2 1  z  z 
1 z z 
1
2 1
GENERAL FORM OF INVERSE MOTIONS
(orientation-reversing isometries)
w  ei z  c
Direct motions - composition of two reflections.
Inverse motions - composition of three reflections.
z z 
Back to the reflection formula w  z  2 1  z  z 
1 z z 
1
2 1
It is particularly useful to consider the case in which z1 , z2 are points on the unit
circle, that is, z1  z2  1, or z1 
1
1
and z2  . The reflection w of z becomes
z1
z2
w  z1  z2  z1 z2 z
As a corollary, if z1  z2  1, the projection of z onto the line through z1, z2 is
zw 1
p
  z  z1  z2  z1 z2 z 
2
2
A PRACTICAL ADVICE
Equations become simpler if we can make the assumption that some
of the complex numbers involved have modulus 1.
Line through z1 , z2 :
z  z1
z  z1

z2  z1 z2  z1
 z  z2  z1   z  z2  z1   z1 z2  z2 z1
We get a simpler form if we can assume z1  z2  1, since the conjugate
function for numbers of modulus one is the same with the inverse:
z  z1 z2 z  z1  z2
Let a  1. Then if we set z1  z2  a, it turns out that the equation
of the tangent line to the unit circle at the point a reduces to:
z  a 2 z  2a
ORTHOCENTERS
The effectiveness of complex numbers in solving problems of triangle geometry
may increase if we assume that the vertices of the triangle ABC under discussion
are points on the unit circle: a  b  c  1.
THEOREM. Let H  h  be the orthocenter of the triangle ABC,
with a  b  c  1. Then h  a  b  c.
Check that h  a  c  b :
1 1

ha
bc
bc
ha
 _______  b c  

1 1
c b
c b
c b

c b
c b
_______
_______
Similarly h  b  c  a and h  c  b  a.
THE FEET OF THE ALTITUDES...
Now that we discussed about orthocenters, it makes sense
to find out the complex numbers corresponding to the feet
of the altitudes of our triangle ABC remember, a  b  c 1 .


1
 z  z1  z2  z1 z2 z  represents the projection of z
2
onto the line through z1and z2 where z1  z2  1.
Recall that
Thus the foot of the altitude from A is
1
1
bc
 a  b  c  bca    a  b  c  
2
2
a 
Similarly, the foot of the altitude from B is
the foot of the altitude from C is
1
ac 
a bc 
 , and
2
b 
1
ab 
a

b

c


.
2
c 
THE NINE  POINT CIRCLE
a  b  c  ORTHOCENTER
abc
 CENTROID ("center of gravity")
3
So far we are aware of h 
What about e :
abc
abc
and g 
1
3
abc
other than being the midpoint of OH ?
2
ab bc ca
,
,
of the sides AB, BC , CA are at a distance
2
2
2
abc
ab
c
1
1 2 from e 
. Indeed, e 

 , etc.
2
2
2
2
* The midpoints
* The feet of the altitudes
1
bc  1 
ac  1 
ab 
a

b

c

,
a

b

c

,
a

b

c


 
 

2
a  2
b  2
c 
are also at a distance 1 2 from e : e 
1
bc 
bc
1
 , etc.
a bc 
 
2
a 
2a
2
We need three more points to complete a beautiful theorem!
* The midpoints of the three segments from the orthocenter to the vertices,
1
1
1
 a  h  ,  b  h  and  c  h  are at a distance 1 2 from e.
2
2
2
Indeed, e 

1
abc 1
  a   a  b  c 
a  h 
2
2
2
a
1
 , etc.
2
2
THEOREM (THE NINE  POINT CIRCLE, OR THE EULER'S CIRCLE)
Given any triangle ABC , the nine points listed below all lie on the same circle
(Euler's Circle) centered at the midpoint between the circumcenter and the
orthocenter of the triangle:
The midpoints of the three sides of the triangle,
The feet of the three altitudes of the triangle, and
The midpoints of the three segments from the orthocenter to the vertices.
INSCRIPTIBLE QUADRILATERALS
Let z1 , z2 , z3 , z4  C represent the vertices of
an inscriptible quadrilateral. Let E1 , E2 , E3 , E4
be the Euler circles of the triangles z2 z3 z4 ,
z1 z3 z4 , z1 z2 z4 and z1 z2 z3 , respectively, with
centers e1 , e2 , e3 , e4 .Then E1 , E2 , E3 , E4 have a
common point. For the proof we may assume,
as usual, that z1  z2  z3  z4  1.
z 2  z3  z 4
z  z3  z 4
, e2  1
,
2
2
z1  z2  z3
z1  z2  z4
e3 
, e4 
.
2
2
z  z 2  z3  z 4
Define e : 1
.
2
e1 
e  e1  e  e2  e  e3  e  e4 
1
 e belongs to each all Euler circles E1 , E2 , E3 , E4 .
2
e1 , e2 , e3 , e4 lie on the same (blue) circle of radius 1 2 centered at e.
This will be the "Euler Circle of the inscriptible quadrilateral z1 z2 z3 z4"
BACK TO PROJECTIONS!!
RECALL: If z1  z2  1, then the projection of z onto the line through z1 , z2 is
1
 z  z1  z2  z1 z2 z 
2
Consider a triangle A1 A2 A3 with A1  z1  , A2  z2  A3  z3  , z1  z2  z3  1,
and let P  z  be an arbitrary point in the plane of the triangle A1 A2 A3 .
Let P1  p1  , P2  p2  , P3  p3  be the projections of P onto the sides A2 A3 , A1 A3 and A1 A2respectively.
p1 
1
 z  z2  z3  z2 z3 z 
2
p2 
1
 z  z1  z3  z1 z3 z 
2
p3 
1
 z  z1  z2  z1 z2 z 
2
SIMSON's THEOREM
The three projections, p1 , p2 , p3 are collinear if and only if z  1, that is,
if and only if P  z  is on the circumcircle of the triangle A1 A2 A3 .
P
R
p2  p1 
1
1
1
 z  z1  z3  z1 z3 z    z  z2  z3  z2 z3 z    z2  z1  z3 z  1
2
2
2
p3  p1 
1
1
1
 z  z1  z2  z1 z2 z    z  z2  z3  z2 z3 z    z3  z1  z2 z  1
2
2
2
O
O
F
___________
p  p1
p  p1
p1 , p2 , p3 collinear  3
 3
p2  p1 p2  p1
 1 1  z

     1
 z  z  z z  1   z3 z1   z2 
 3 1 2
 z2  z1  z3 z  1  1  1   z  1



 z2 z1   z3 

 z3  z1  z2 z  1   z3  z1  z2 z  1
 z2  z1  z3 z  1  z2  z1  z3 z  1


z2 z  1 z  z2
2


  z2  z3  z  1  0
z3 z  1 z  z3
z 1
SIMSON's THEOREM  A GENERALIZATION
With the previous notation in place, we want to characterize the set of all
points P  z  such that the (oriented) area of the triangle P1 P2 P3 determined
by the projections of P onto the sides of A1 A2 A3 is a given constant.
Area  p1 p2 p3  
1
Im  p3  p1  p2  p1  
2
1
 p3  p1  p2  p1    p3  p1  p2  p1  
4i
1
1
RECALL: p3  p1   z3  z1  z2 z  1 and p2  p1   z2  z1  z3 z  1
2
2
Area  p1 p2 p3  
Area  p1 p2 p3  

1
 z3  z1  z2 z  1 z2  z1  z3 z  1   z3  z1  z2 z  1 z2  z1  z3 z  1 
16i 

  1 1  z
 1 1  z

1 
 z3  z1  z2 z  1      1       1  z2  z1  z3 z  1 
16i 

 z2 z1   z3   z3 z1   z2 
Area  p1 p2 p3 
z2  z1  z1  z3  z3  z2 

1  z 2 

16iz1 z2 z3


AN INTERESTING FUNCTION
Let U   z  C : z  1. DEFINE f : U 3  C , f  z1 , z2 , z3  
 z2  z1  z1  z3  z3  z2 
z1 z2 z3
FACT 1. f  z1 , z2 , z3  is PURELY IMAGINARY for all z1 , z2 , z3 U .
_________________
PROOF. f  z1 , z2 , z3  

 z2  z1  z1  z3  z3  z2 
 z1  z2  z3  z1  z2  z3    f
z1 z2 z3
z1 z2 z3
 1 1  1 1  1 1 
      
z
z1   z1 z3   z3 z2 
 2
1
z1 z2 z3
 z1 , z2 , z3 
FACT 2. f  z1 , z2 , z3   4  Area  z1 z2 z3 
PROOF. f  z1 , z2 , z3   z1  z2 z3  z1 z2  z3  4  R  Area  z1 z2 z3   4  Area  z1 z2 z3 


inv 
FACT 3. f  z1 , z2 , z3  is antisymmetric: f z 1 , z  2 , z 3   1
f  z1 , z2 , z3 
FACT 4. f 1, i, 1  4i  4i  Area 1, i, 1
CONCLUSION: f  z1 , z2 , z3   4iArea  z1 z2 z3 
Area  p1 p2 p3 
z2  z1  z1  z3  z3  z2 

2


1 z 
16iz1 z2 z3


&
 z2  z1  z1  z3  z3  z2   4iArea
z1 z2 z3
 z1 z2 z3 

Area  p1 p2 p3  
1 z
4
2
 Area  z1 z2 z3 
EXAMPLE: In the special case z  0, p1 , p2 , p3 are the midpoints of the sides
of z1 z2 z3 and Area  p1 p2 p3   Area  z1 z2 z3  4.
EULER CIRCLES AND SIMSON LINES
RECALL: z1 z2 z3 z4 inscriptible quadrilateral,
E1 , E2 , E3 , E4  Euler circles of the triangles
z2 z3 z4 , z1 z3 z4 , z1 z2 z4 and z1 z2 z3 , e1 , e2 , e3 , e4
their centers. Then e   z1  z2  z3  z4  2
is the center of the Euler circle or z1 z2 z3 z4 ,
of radius 1 2, containing e1 , e2 , e3 , e4 .
CONNECTION WITH SIMSON LINES:
If z1 z2 z3 z4 is an inscriptible quadrilateral,
the 4 Simson lines: of z1 with respect to
z2 z3 z4 ,of z2 with respect to z1 z3 z4 ,of z3 with
respect to z1 z2 z4 and of z4 with respect to
z1 z2 z3 , are concurrent, all passing through
the center e of the Euler circle of z1 z2 z3 z4 .
AN EQUATION FOR THE SIMSON LINE
Let z be on the circumcircle of the triangle z1 z2 z3 . The three projections, p1 , p2 , p3 of z onto
the lines z2 z3 , z1 z3 , z1 z2 are collinear, belonging to the Simson line l of z with respect to z1 z2 z3 .
The equation of l is
t  p1
t  p1

 t  p2  p1   t  p2  p1   p1 p2  p1 p2
p2  p1 p2  p1
 p  p1 
p1 p2  p1 p2
tt  2

C
,
where
C:=

p

p
p2  p1
1 
 2
RECALL: p2  p1 
p2  p1 
z
1
1
1
 z2  z1  z3 z  1   z2  z1   3  1   z2  z1  z3  z 
2
2
 z
 2z

1
1  1 1  z
1
 z2  z1  z3 z  1       1 
 z2  z1  z3  z 
2
2  z2 z1  z3  2 z1 z2 z3
 p2  p1 
z1 z2 z3
Thus t  t 
C
C  tt
z
 p2  p1 
Set t  p1 to determine C.
z  z1  z2  z3 =1  The equation of the Simson line of z with respect to z1 z2 z3 is of the form
tt
z z 
1
Set t  p1   z  z2  z3  2 3 
2
z 
z1 z2 z3
C
z
z z  zz z 
z z 
1
We get C   z  z2  z3  2 3   1 2 3  z  z2  z3  2 3  
2
z 
2z 
z 
zz z
1
  z  z1  z2  z3   1 2 3  z  z1  z2  z3 
2
2z
 The equation of the Simson line of z with respect to z1 z2 z3 is 
z  z1  z2  z3

Passes
through
t

zz z
z  z1  z2  z3  z1 z2 z3  z  z1  z2  z3

2
t 1 2 3 t 



z
2
2
 z 

CONCLUDING THEOREM ON SIMSON LINES AND EULER CIRCLES:
If z1 z2 z3 z4 is an inscriptible quadrilateral, the 4 Simson lines: of z1 with respect to z2 z3 z4 ,
of z2 with respect to z1 z3 z4 ,of z3 with respect to z1 z2 z4 and of z4 with respect to z1 z2 z3 , are
concurrent, all passing through the center e 
z1  z2  z3  z4
of the Euler circle of z1 z2 z3 z4 .
2
A PUTNAM EXAM PROBLEM INVOLVING ROTATIONS
The 65th William Lowell Putnam Mathematical Competition, 2004.
PROBLEM B4.
Let n be a positive integer, n  2, and put   2 / n. Define points Pk  ( k , 0)
in the xy  plane, for k  1, 2,..., n. Let Rk be the map that rotates the plane
counterclockwise by the angle  about the point Pk . Let R denote the map
obtained by applying, in order, R1 , then R2 ,..., then Rn . For an arbitrary point
( x, y ), find, and simplify, the coordinates of R( x, y ).
Since 2 n  2 n  ...  2 n  2 , we can say, from our general knowledge about
n summands
compositions of rotations, that the transformation relating R  x, y  to  x, y  must be
a translation!
A "VISUAL" PROOF...
1
z0  x0  iy0
2
4
3
R1  z0 
R2  z1 
R3  z2 
z1
z2
z3
...
...
n 1
n
Rn  zn 1 
zn
The point z0  x0  iy0 is chosen such that the segment from z0 to z1  R1  z0  has length 1, is horizontal and
in the lower half plane. Thus, the x  coordinates of z0 , z1 , z2 ,...,z n are half-integers: 1 2,3 2,...,  2n  1 2,


respectively, and they have the same y  coordinate,  1  2 tan  .
n

Therefore in the above particular picture, R  z0   z0  n. Since we know that R is a translation,
it turns out that R  z   z  n for all z C .
A PROOF BASED ON COMPLEX NUMBERS
Let   e2 i n  cos  2 n   i sin  2 n  . The multiplication by  signifies
a counterclockwise rotation with an angle of 2 n .
z1  1    z  1  z1   z  1    1
z2   z1  1     2  z2   2 z   1    1  1     2
z3   z2  1     3  z3   3 z   2 1    1   1     2  1     3
zn   n z   n 1 1    1   n  2 1     2  ...   1      n  1  1     n
n
zn  z  1     k n  k
k 1
n
The translation vector will be t  1     k
k 1
nk
n
 1     k  k , since  n  1.
k 1
n
We use the identity  kx k 


x n nx 2  nx  x  x
 x  1
k 1
2
The translation vector will be:
n
t  1     k
k
 1   
n 2  n 1   1   1

k 1
 1   
n 2  n 1

1

1
2
1

1
2
 1   

 
since  -1
n  n
1   
2
n

1
n
This concludes the proof by using complex rotations. The geometrical proof seems,
at least in the case of this particular problem, easier. However, if we insist on dealing
with complex numbers we will be rewarded by an interesting generalization of this
nice Putnam problem.
A GENERALIZATION OF THE PUTNAM B4 (2004)
Let n  2 and let z0 , z1 ,..., zn 1  C . For every k  0,1,..., n  1 , the composition
of the rotations with 2k n around z0 , z1 ,..., zn 1  in this order  is a translation
with a vector which we shall call tk . This is because the sum of the angles of rotation
is 2k  2 Z .
Which kind of relation is there between the ordered n  tuple  z0 , z1 ,..., zn 1  of centers of rotation
on one hand, and the ordered n  tuple  t0 , t1,..., tn 1  of translation vectors, on the other hand?
Let k 0,1,..., n 1 and let  :  k  e2k i n . Let z C be arbitrary.
ROTATION AROUND z0 , z
w0
Then w0  z0    z  z0   w0   z  1    z0
ROTATION AROUND z1 , w0
w1
Then w1  z1    w0  z1   w1   w0  1    z1  w1   2 z   1    z0  1    z1
ROTATION AROUND z2 , w1
w2
w2   3 z   2 1    z0   1    z1  1    z2 , etc.
WE EVENTUALLY GET...
wn 1   n z   n 1 1    z0   n  2 1    z1   n 3 1    z2  ...   1    zn 2  1    zn 1
Or, since  n  1,
wn 1  z   n 1 1    z0   n  2 1    z1   n 3 1    z2  ...   1    zn  2  1    zn 1
That is,
tk   n 1 1    z0   n  2 1    z1   n 3 1    z2  ...   1    zn 2  1    zn 1
Since    k  e2k i n , we get:

tk  1  
n 1
k
 z  
k n  r 1
r
r 0
Since  n  1,


tk  1   
k
k
n 1
z 
r 0
r
 kr

 
k
n 1
 z e
1
r 0
r
2 k i
n


   k  1 zˆk
We have thus proved the following
THEOREM. Let n  2 and let z0 , z1 ,..., zn 1  C . Then for every k  0,1,..., n  1 , the
composition of the rotations with 2k n around z0 , z1 ,..., zn 1  in this order  is a translation


with a vector tk    k  1 zˆk , where  zˆ0 , zˆ1 ,..., zˆn 1   C n represents the discrete Fourier
n 1
transform of  z0 , z1 ,..., zn 1   C , that is, zˆk   zr e
n
2 k i
n
, for k  0,..., n  1.
r 0
USEFUL REFERENCES:
1. I.M. YAGLOM, Complex Numbers in Geometry, Academic Press, 1968.
2. LIANG-SHIN HAN, Complex Numbers & Geometry, MAA, 1994.