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.