Geometry with Complex Numbers Mihai Caragiu Ohio Northern University Abstract: In the last three...
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Transcript of Geometry with Complex Numbers Mihai Caragiu Ohio Northern University Abstract: In the last three...
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
2
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 +1=0
x
IMAGINARY UNIT
2 1i
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, ,
Oxford University Press, 1958, Ch.7)
An Introduction to Mathematics
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
1 1 2 2 1 2 1 2
1 1 2 2 1 2 1 2 1 2 2 1
= | ,x iy x y
x iy x iy x x i y y
x iy x iy x x y y i x y x y
C R
2 2
For define
Re( ) , Im( ) , ,
z x iy
z x z y z x iy z x y z z
so that...
1 1 22
2 2
Re( ) , Im( ) , 2 2
z z zz z z zz z
i z z
FIRST CONTACT: RECTANGULAR FORM
REACTANGULAR REPRESENTATION:
GEOMETRICAL CONNECTIONS
may be seen asz x iy The point , in the plane x y
The plane vector , x y
1z
2z 2 1 z z
Thus the transformation given by
is a TRANSLATION with the vector corresponding
to the complex number .
z w w z k
k
reflection about the axis.w z x reflection about the origin.w z
reflection about the axis.w z y
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:
1 2
1 2
1 1
2 1 2 2 1 1 1 2 1 2 1 2 2 1
2 22
OP OPArea OP P
P zz z x iy x iy x x y y i x y x y
P z
1 1 2 2CONSEQUENCES: Let A ,B ,..., , ,...a b P z P z
* The general eq. of a line perpendicular to is Re constant. AB z b a
* The general eq. of a line parallel to is Im constant. AB z b a
1 2 3 4 4 3 2 1* The segments and are perpedicular Re z =0. PP P P z z z4 3
1 2 3 42 1
z* and are perpedicular purely imaginary.
z
PP P Pz z
4 31 2 3 4
2 1
z* and are parallel purely real.
z
PP P Pz z
POLAR AND EXPONENTIAL FORMS
cos sin 0, 2iz r i re r R Z
"MULTIPLICATIVE FRIENDLY":
1 2
1 2 1 2 1 2 1 2 1 2cos sin . iz z r r i r r e
SPECIAL CASE
CONSIDER COMPLEX NUMBERS WITH u 1, cos sin . i
u u i e
iii i
i
u ew uz e re re
z re
THE TRANSFORMATION
IS A ROTATION WITH ANGLE ABOUT THE ORIGIN w uz
"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 3Let = cos sin .
2 2 3 3
i i
3 21, 1. Consider the equilateral
triangle with vertices A 0 ,B 1 , and
an arbitrary point P .
C
z
If we rotate P with about the origin, we get P . Then the sides of the triangle 3
with vertices , , are identical with the distances z , 1 and from
to the vertices 0,1, of the equila
z z P PC
z z zz zteral triangle we considered initially.
221 and 1 1 z z z z z z z z z
1 2 3 1 2 3 1 2 3 1 2 3
1 3 1 2
Let , , , , , be complex numbers. Then the triangles z and
are similar if the way of obtaining the complex segment z z from z z is the same
as the way of obtaining the complex s
z z z w w w z z w w w
1 3 1 2egment from .w w w w
1 3 1 2
1 1 1 3
1 2
For example, if can be obtained from by a rotation with an angle of around5
followed by a dilation by a factor of 2 with respect to , then can be obtained
from by a rotation
z z z z
z z w w
w w1
1
with an angle of around followed by a dilation by a 5
factor of 2 with respect to .
w
w
SIMILARITY - a problem of rotations and homotheties...
3 1 3 11 2 3 1 2 3
2 1 2 1
z zIf z z z ~ and if , then z z
w ww w w a aw w
3 1 3 11 2 3 1 2 3
2 1 2 1
z zz z z ~ if and only if z z
w ww w w w w
GENERAL ROTATIONS IN THE PLANE
0 0 0
The transformation formula for a counterclockwise rotation by an
angle about a point is . iz w z e z z
01 i iw e z e z
GENERAL FORM OF DIRECT MOTIONS
(orientation-preserving isometries)
a translation, if 2 , or otherwise
is a rotation of angle around 1
i
i
w e z c c
e
Z
1 2
1 2
1 2 2 1 1 2
3 1 21 2
1 2
, if 2 R
(translation) if 2
COMPOSITIONS: R R ; k
k k
k k k k k k
R zz R z
T
z T z T R z R z
T T T T T
Z
Z
Let , , . Let . We need to find the 1 2 1 2
reflection of about the line through and . 1 2
z z z z z
w z z z
C C
REFLECTIONS: orientation-reversing isometries
1 2 1Thus 1 2 1 2 1 12 1 2 1
z z z zw z z z z z z zz z z z
1Let = . Then and .1 2 1 1 2 12 1
z zz z z z w z z zz z
2 11 12 1
z zw z z zz z
GENERAL FORM OF INVERSE MOTIONS
(orientation-reversing isometries)
iw e z cDirect motions - composition of two reflections.
Inverse motions - composition of three reflections.
2 1Back to the reflection formula 1 12 1
z zw z z zz z
1 2
1 2 1 21 2
It is particularly useful to consider the case in which , are points on the unit
1 1circle, that is, 1, or and . The reflection of becomes
z z
z z z z w zz z
1 2 1 2 w z z z z z
1 2 1 2As a corollary, if 1, the projection of onto the line through , is z z z z z
1 2 1 2
1
2 2
z wp z z z z z z
A PRACTICAL ADVICE
1 11 2
2 1 2 1
Line through , : z z z z
z zz z z z
2 1 2 1 1 2 2 1z z z z z z z z z z
1 2We get a simpler form if we can assume z 1, since the conjugate
function for numbers of modulus one is the same with the inverse:
z
1 2 1 2 z z z z z z
1 2Let 1. Then if we set , it turns out that the equation
of the tangent line to the unit circle at the point reduces to:
a z z a
a
2 2 z a z a
Equations become simpler if we can make the assumption that some
of the complex numbers involved have modulus 1.
ORTHOCENTERSThe 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: 1. a b c
THEOREM. Let be the orthocenter of the triangle ABC,
with 1. Then .
H h
a b c h a b c
Check that : h a c b
_______ _______
_______
1 1
1 1
h a b c b c h ab cc b c b c bc b
c b
Similarly and . h b c a h c b a
THE FEET OF THE ALTITUDES...
Now that we discussed about orthocenters, it makes senseto find out the complex numbers corresponding to the feet
of the altitudes of our triangle ABC remember, 1 . a b c
1 2 1 2
1 2 1 2
1Recall that represents the projection of
2onto the line through and where 1.
z z z z z z z
z z z z
Thus the foot of the altitude from A is
1 1
2 2
bca b c bca a b c
a
1Similarly, the foot of the altitude from B is , and
2
aca b c
b
1the foot of the altitude from C is .
2
aba b c
c
ORTHOCENTER a b c
THE NINE POINT CIRCLE
CENTROID ("center of gravity")3
a b c
So far we are aware of and 1 3
What about : other than being the midpoint of ?2
a b c a b ch g
a b ce OH
* The midpoints , , of the sides , , are at a distance 2 2 2
11 2 from . Indeed, , etc.
2 2 2 2
a b b c c aAB BC CA
a b c a b ce e
1 1 1* The feet of the altitudes , ,
2 2 2
1 1are also at a distance 1 2 from : , etc.
2 2 2
bc ac aba b c a b c a b c
a b c
bc bce e a b c
a a
* The midpoints of the three segments from the orthocenter to the vertices,
1 1 1 , and are at a distance 1 2 from .
2 2 2 a h b h c h e
We need three more points to complete a beautiful theorem!
1 1Indeed,
2 2 2
1, .
2 2
a b ce a h a a b c
aetc
THEOREM (THE NINE POINT CIRCLE, OR THE EULER'S CIRCLE)
Given any triangle , the nine points listed below all lie on the same circle
(Euler's Circle) centered at the midpoint between the circumcenter
ABC
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.
1 2 3 4
1 2 3 4
2 3 4
1 3 4 1 2 4 1 2 3
1 2 3 4
Let , , , represent the vertices of
an inscriptible quadrilateral. Let , , ,
be the Euler circles of the triangles ,
, and , respectively, with
centers , , ,
z z z z
E E E E
z z z
z z z z z z z z z
e e e e
C
1 2 3 4
1 2 3 4
.Then , , , have a
common point. For the proof we may assume,
as usual, that 1.
E E E E
z z z z
2 3 4 1 3 41 2
1 2 31 2 43 4
1 2 3 4
, ,2 2
, .2 2
Define : .2
z z z z z ze e
z z zz z ze e
z z z ze
1 2 3 4 1 2 3 4
1 belongs to each all Euler circles , , , .
2e e e e e e e e e E E E E
INSCRIPTIBLE QUADRILATERALS
1 2 3 4
1 2 3 4
, , , lie on the same (blue) circle of radius 1 2 centered at .
This will be the "Euler Circle of the inscriptible quadrilateral "
e e e e e
z z z z
1 2 1 2
1 2 1 2
RECALL: If 1, then the projection of onto the line through , is
1
2
z z z z z
z z z z z z
BACK TO PROJECTIONS!!
1 2 3 1 1 2 2 3 3 1 2 3
1 2 3
Consider a triangle with , , 1,
and let P z be an arbitrary point in the plane of the triangle .
A A A A z A z A z z z z
A A A
1 1 2 2 3 3 2 3 1 3 1 2Let , , be the projections of P onto the sides , and respectively. P p P p P p A A A A A A
1 2 3 2 3
1
2 p z z z z z z
2 1 3 1 3
1
2 p z z z z z z
3 1 2 1 2
1
2 p z z z z z z
SIMSON's THEOREM
1 2 3
1 2 3
The three projections, , , are collinear if and only if 1, that is,
if and only if is on the circumcircle of the triangle .
p p p z
P z A A A
2 1 1 3 1 3 2 3 2 3 2 1 3
1 1 11
2 2 2 p p z z z z z z z z z z z z z z z z
3 1 1 2 1 2 2 3 2 3 3 1 2
1 1 11
2 2 2 p p z z z z z z z z z z z z z z z z
___________
3 1 3 11 2 3
2 1 2 1
, , collinear
p p p pp p p
p p p p
3 1 2 3 1 2
2 1 3 2 1 3
1 1
1 1
z z z z z z z z
z z z z z z z z
3 1 2 3 1 2
2 1 3
2 1 3
1 11
1
1 1 11
zz z z z z z z
z z z z zz z z
2 2
3 3
1
1
z z z z
z z z z 2
2 3 1 0 z z z
z 1
P
R
O
O
F
SIMSON's THEOREM A GENERALIZATION
1 2 3
1 2 3
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 determined
by the projections of onto the sides of i
PP P
P A A A s a given constant.
1 2 3 3 1 2 1
1Area Im
2 p p p p p p p
1 2 3 3 1 2 1 3 1 2 1
1Area
4 p p p p p p p p p p p
i
3 1 3 1 2 2 1 2 1 3
1 1RECALL: 1 and 1
2 2 p p z z z z p p z z z z
1 2 3 3 1 2 2 1 3 3 1 2 2 1 3
1Area 1 1 1 1
16 p p p z z z z z z z z z z z z z z z z
i
3 1 2 2 1 32 1 3 3 1 2
1 1 1 1 11 1 1 1
16
z zz z z z z z z z
i z z z z z z
22 1 1 3 3 21 2 3
1 2 3
Area 116
z z z z z z
p p p ziz z z
AN INTERESTING FUNCTION
2 1 1 3 3 231 2 3
1 2 3
Let : 1 . DEFINE : , , ,z z z z z z
U z z f U f z z zz z z
C C
1 2 3 1 2 3
_________________2 1 1 3 3 2 2 1 1 3 3 2
1 2 31 2 3
1 2 3
1 2 3 1 2 31 2 3
1 2 3
FACT 1. , , is PURELY IMAGINARY for all , , .
1 1 1 1 1 1
PROOF. , ,1
, ,
f z z z z z z U
z z z z z z z z z z z zf z z z
z z zz z z
z z z z z zf z z z
z z z
1 2 3 1 2 3
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
FACT 2. , , 4 Area
PROOF. , , 4 Area 4 Area
f z z z z z z
f z z z z z z z z z R z z z z z z
1 2 3 1 2 31 2 3FACT 3. , , is antisymmetric: , , 1 , ,
invf z z z f z z z f z z z
FACT 4. 1, , 1 4 4 Area 1, , 1 f i i i i
1 2 3 1 2 3CONCLUSION: , , 4 Areaf z z z i z z z
22 1 1 3 3 21 2 3
1 2 3
2 1 1 3 3 21 2 3
1 2 3
Area 116
&
4 Area
z z z z z zp p p z
iz z z
z z z z z zi z z z
z z z
2
1 2 3 1 2 3
1Area Area
4
zp p p z z z
1 2 3
1 2 3 1 2 3 1 2 3
EXAMPLE: In the special case 0, , , are the midpoints of the sides
of and Area Area 4.
z p p p
z z z p p p z z z
EULER CIRCLES AND SIMSON LINES
1 2 3 4
1 2 3 4
2 3 4 1 3 4 1 2 4 1 2 3 1 2 3 4
1 2 3 4
RECALL: inscriptible quadrilateral,
, , , Euler circles of the triangles
, , and , , , ,
their centers. Then 2
is the center of the Euler circle o
z z z z
E E E E
z z z z z z z z z z z z e e e e
e z z z z
1 2 3 4
1 2 3 4
r ,
of radius 1 2, containing , , , .
z z z z
e e e e
1 2 3 4
1
2 3 4 2 1 3 4 3
1 2 4 4
CONNECTION WITH SIMSON LINES:
If is an inscriptible quadrilateral,
the 4 Simson lines: of with respect to
,of with respect to ,of with
respect to and of with resp
z z z z
z
z z z z z z z z
z z z z
1 2 3
1 2 3 4
ect to
, are concurrent, all passing through
the center e of the Euler circle of .
z z z
z z z z
AN EQUATION FOR THE SIMSON LINE
1 2 3 1 2 3
2 3 1 3 1 2 1 2 3
Let be on the circumcircle of the triangle . The three projections, , , of onto
the lines , , are collinear, belonging to the Simson line of with respect to .
z z z z p p p z
z z z z z z l z z z z
1 12 1 2 1 1 2 1 2
2 1 2 1
The equation of is t p t p
l t p p t p p p p p pp p p p
2 1 1 2 1 2
2 1 2 1
, where C:=p p p p p p
t t Cp p p p
32 1 2 1 3 2 1 2 1 3
1 1 1RECALL: 1 1
2 2 2
zp p z z z z z z z z z z
z z
2 1 2 1 3 2 1 32 1 3 1 2 3
1 1 1 1 11 1
2 2 2
zp p z z z z z z z z
z z z z z z
1 2 32 1
2 1
Thus z z zp p
t t C t t Cp p z
1Set to determine C.t p
1 2 3 1 2 3
1 2 3
=1 The equation of the Simson line of with respect to is of the form
z z z z z z z z
z z zt t C
z
2 31 2 3
1Set
2
z zt p z z z
z
2 3 1 2 3 2 32 3 2 3
1 2 31 2 3 1 2 3
1We get C
2 2
1
2 2
z z z z z z zz z z z z z
z z z
z z zz z z z z z z z
z
1 2 3
1 2 3 1 2 3 1 2 3 1 2 3
The equation of the Simson line of with respect to is
2 2
z z z z
z z z z z z z z z z z z z zt t
z z
1 2 3Passes through 2
z z z zt
1 2 3 4 1 2 3 4
2 1 3 4 3
CONCLUDING THEOREM ON SIMSON LINES AND EULER CIRCLES:
If is an inscriptible quadrilateral, the 4 Simson lines: of with respect to ,
of with respect to ,of with respect to
z z z z z z z z
z z z z z z1 2 4 4 1 2 3
1 2 3 41 2 3 4
and of with respect to , are
concurrent, all passing through the center of the Euler circle of .2
z z z z z z
z z z ze z z z z
A PUTNAM EXAM PROBLEM INVOLVING ROTATIONS
Since 2 2 ... 2 2 , we can say, from our general knowledge about
compositions of rotations, that the transformation relating , to , must be
a !
n summands
n n n
R x y x y
translation
The 65th William Lowell Putnam Mathematical Competition, 2004.
PROBLEM B4.
Let be a positive integer, 2, and put 2 / . Define points ( , 0)
in the plane, for 1, 2,..., . Let be the mapk
k
n n n P k
xy k n R
1 2
that rotates the plane
counterclockwise by the angle about the point . Let denote the map
obtained by applying, in order, , then ,..., then . For an arbitrary point
( , ), find, and simp
k
n
P R
R R R
x y
lify, the coordinates of ( , ).R x y
A "VISUAL" PROOF...
0 0 0 z x iy 1
1 0 z
R z 2
2 1 z
R z 3
3 2 z
R z ... 1
n
n n
z
R z
1 2 3 4 ... 1n n
0 0 0 0 1 1 0
0 1 2 n
The point is chosen such that the segment from to has length 1, is horizontal and
in the lower half plane. Thus, the coordinates of , , ,...,z are half-integers: 1 2 ,3 2,...,
z x iy z z R z
x z z z
2 1 2,
respectively, and they have the same coordinate, 1 2 tan .
n
yn
0 0Therefore in the above particular picture, . Since we know that R is a translation,
it turns out that for all .
R z z n
R z z n z
C
A PROOF BASED ON COMPLEX NUMBERS 2Let cos 2 sin 2 . The multiplication by signifies
a counterclockwise rotation with an angle of 2 .
i ne n i n
n
1 11 1 1 1 z z z z
22 1 21 2 1 1 1 2 z z z z
3 23 2 31 3 1 1 1 2 1 3 z z z z
1 21 1 1 2 ... 1 1 1 n n nnz z n n
1
1
n
n kn
k
z z k
1 1
The translation vector will be 1 1 , since 1.
n n
n k k n
k k
t k k
2
21
We use the identity 1
nnk
k
x nx nx x xkx
x
2 1 1 1-1
211
The translation vector will be:
1 1 since 11
n nk
k
n nt k
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.
2 1
2 211 1
11
n n n nn
0 1 1
0 1 1
Let 2 and let , ,..., . For every 0,1,..., 1 , the composition
of the rotations with 2 around , ,..., in this order is a translation
with a vector which we shall call . This is b
n
n
k
n z z z k n
k n z z z
t
C
ecause the sum of the angles of rotation
is 2 2 .k Z
0 1 1
0 1 1
Which kind of relation is there between the ordered tuple , ,..., of centers of rotation
on one hand, and the ordered tuple , ,..., of translation vectors, on the other hand?
n
n
n z z z
n t t t
2Let 0,1,..., 1 and let : . Let be arbitrary.k k i nk n e z C
0 0
0 0 0 0 0
ROTATION AROUND ,
Then 1
z z w
w z z z w z z
1 0 1
21 1 0 1 1 0 1 1 0 1
ROTATION AROUND ,
Then 1 1 1
z w w
w z w z w w z w z z z
2 1 2
3 22 0 1 2
ROTATION AROUND ,
1 1 1 , .
z w w
w z z z z etc
A GENERALIZATION OF THE PUTNAM B4 (2004)
1 2 31 0 1 2 2 11 1 1 ... 1 1
n n n nn n nw z z z z z z
WE EVENTUALLY GET...
Or, since 1, n
1 2 31 0 1 2 2 11 1 1 ... 1 1
n n nn n nw z z z z z z
That is,
1 2 30 1 2 2 11 1 1 ... 1 1n n n
k n nt z z z z z
1
1
0
1n
k n rkk r
r
t z
2Since , we get:k k i ne
Since 1,n
21 1
0 0
ˆ1 1 1k in n
k k kr k knk r r k
r r
t z z e z
We have thus proved the following
0 1 1
0 1 1
THEOREM. Let 2 and let , ,..., . Then for every 0,1,..., 1 , the
composition of the rotations with 2 around , ,..., in this order is a translation
ˆwith a vector 1 , where
n
n
kk k
n z z z k n
k n z z z
t z
C
0 1 1
21
0 1 10
ˆ ˆ ˆ , ,..., represents the discrete Fourier
ˆtransform of , ,..., , that is, , for 0,..., 1.
nn
k inn n
n k rr
z z z
z z z z z e k n
C
C
USEFUL REFERENCES:
1. I.M. YAGLOM, Complex Numbers in Geometry, Academic Press, 1968.
2. LIANG-SHIN HAN, Complex Numbers & Geometry, MAA, 1994.