3 Analytic Geometry
Geometry in the style of Euclid and Hilbert is synthetic: axiomatic, without co-ordinates or explicit
numerical measures of angle, length, area or volume. By contrast, the modern practice of geometry
is typically analytic: reliant on algebra and co-ordinates (including vectors). The critical development
came in the early 1600s courtesy of Ren
´
e Descartes and Pierre de Fermat: the axis as a fixed reference
ruler against which objects can be measured using co-ordinates.
3.1 The Cartesian Co-ordinate System
Since Cartesian geometry (Descartes’ geometry) should be familiar, we merely sketch the core ideas.
• Perpendicular axes meet at the origin O.
• The co-ordinates of a point are measured by projecting onto the axes;
since these are real numbers we denote the set of these
R
2
=
(x, y) : x, y ∈ R
In the picture, P has co-ordinates ( 1, 2); we usually write P = (1, 2).
• Algebra is introduced via addition and scalar multiplication
−2
−1
1
2
3
y
−2 −1 1 2 3
x
P
P + Q = (p
1
, p
2
) + (q
1
, q
2
) = (p
1
+ q
1
, p
2
+ q
2
) λP = (λp
1
, λp
2
)
• The length of a segment uses Pythagoras’ Theorem
d(P, Q) =
|
PQ
|
=
q
(q
1
− p
1
)
2
+ (q
2
− p
2
)
2
In the picture,
|
OP
|
=
√
1
2
+ 2
2
=
√
5. As in Section 2.5, segments are congruent if and only if
they have the same length.
• Curves are defined using equations. E.g., x
2
+ y
2
= 1 describes a circle.
Analytic geometry was originally conceived as a computational toolkit built on top of Euclid. Math-
ematicians at first felt the need to justify analytic arguments synthetically lest no-one believe their
work.
15
Synthetic geometry is not without its benefits, but its study has increasingly become a fringe
activity; co-ordinates are just too useful to ignore. We may therefore assume anything from Euclid
and mix strategies as appropriate. To see this at work, consider a simple result.
Lemma 3.1. Non-collinear points O = (0, 0), A = (x, y), B = (v, w) and
C := (x + v, y + w), form a parallelogram OACB.
Proof. Opposite sides have the same length (
|
BC
|
=
p
x
2
+ y
2
=
|
OA
|
, etc.)
and are thus congruent. SAS shows △OAC
∼
=
△CBO. Euclid’s discussion
of alternate angles (pages 10–11) forces opposite sides to be parallel.
O
A
B
C
Lemma 3.1 is essentially vector addition: feel free to use such notation/language if you prefer.
15
An attitude which persisted for some time: the presentation in Issac Newton’s groundbreaking Principia (1687) was
largely synthetic, even though his private derivations made extensive use of co-ordinates and algebra.
35
Lemma 3.2. The points X
t
on the line
←→
PQ are in 1–1 correspondence with the real numbers via
X
t
= P + t(Q − P) = (1 − t)P + tQ
Moreover, d(P, X
t
) =
|
t
||
PQ
|
so that t measures the (signed) distance along the line.
The proof is an exercise. As an example of how easy it can be to work in analytic geometry, we apply
the Lemma to re-establish a famous result (compare Exercise 2.5.8c where we used Ceva’s Theorem!).
Theorem 3.3. The medians of a triangle meet at a point 2/3 of the way along each median.
Proof. Given △ABC, label the midpoints of each side as shown. By Lemma 3.2, these are
M =
1
2
(B + C), N =
1
2
(A + B), P =
1
2
(A + C)
The point
2
3
of the way along median AM is then
G := A +
2
3
(M − A) = A +
2
3
(B + C −2A) =
1
3
(A + B + C)
A
C
B
M
P
N
G
By symmetry (check directly if you like!), G is also
2
3
of the way along the other two medians.
Our proof relied on adding points as abstract objects, though we could instead have expressed
A, B, C, . . . in co-ordinates. Exercise 2 does exactly this in an approach that illustrates one of the
biggest advantages of analytic geometry: the freedom to choose axes and co-ordinates so as to make
calculations simple. This is essentially Euclid’s superposition principle or Hilbert’s congruence in
disguise: we’ll make this correspondence rigorous in Section 3.3 when we discuss isometries.
Exercises 3.1. 1. By completing the square, identify the curve described by the equation
x
2
+ y
2
−4x + 2y = 10
2. (a) Perform a pure co-ordinate proof of Theorem 3.3. For simplicity, arrange the triangle so
that A = (0, 0) is the origin, and B points along the positive x-axis.
(b) Descartes and Fermat did not have a fixed perpendicular second axis! Their approach was
equivalent to choosing a second axis oriented to make the problem as simple as possible.
Given △ABC, choose axes pointing along AB and AC. Describe the co-ordinates of B and
C with respect to such axes. Now give an even simpler proof of the centroid theorem (3.3).
3. Prove Lemma 3.2.
4. A parabola is a curve whose points are equidistant from a fixed point F (the focus) and a fixed
line ℓ (the directrix).
(a) Choose axes as shown in the picture so that F = (0, a) and ℓ has
equation y = −a. Find the equation of the parabola.
(b) Now let e be a positive constant (= 0, 1). What happens if ℓ has
equation y = −
a
e
and
|
PF
|
|
Fℓ
|
= e? What happens in the limit e → 0
+
?
y
x
a
ℓ
F
P
36
3.2 Angles and Trigonometry
Angles are defined differently to Section 2.5, though the approach should feel familiar.
Definition 3.4. Suppose A, B, C are distinct points in the plane. Take
any circular arc centered at A and define the radian measure
∡ABC :=
arc-length
radius
∈ [0, 2π)
where arc-length is measured counter-clockwise from
−→
BA to
−→
BC.
θ
2π − θ
B
A
C
Since arc-length scales with radius, the definition is independent of the radius of the circular arc. It
is important to appreciate the difference between angle measures in our two geometries.
Euclidean geometry All angles < 180°. Reversed legs ⇝ congruent angles and same degree measure:
∠CBA
∼
=
∠ABC and m∠CBA = m∠ABC
Analytic geometry Reflex angles exist (≥ π). Reversed legs ⇝ different radian measures:
∡CBA = 2π −θ = θ = ∡ABC (unless a straight edge)
However, since one arrangement of legs produces a measure ≤ π,
∠XYZ
∼
=
∠ABC ⇐⇒ ∡XYZ = ∡ABC or ∡CBA (= 0, π)
As such, we label angles in a triangle by their measures (degree or radian
< π). Standard convention is shown: (A, a, α) ↭ (point, length, angle).
α
β
γ
a
b
c
A
B
C
Definition 3.5 (Trigonometric Functions). Let O be the origin and I = (1, 0).
Let P = (x, y) lie on a circle of radius r and θ = ∡IOP. We define:
cos θ :=
x
r
sin θ :=
y
r
tan θ :=
y
x
(x = 0)
AAA similarity (Thm. 2.42) says these are well-defined, independent of r.
x
y
r
θ
P
O
I
Example 3.6. Basic trig identities should be obvious from the picture: e.g.,
cos
2
θ + sin
2
θ = 1 (Pythagoras!) and sin θ = cos(
π
2
−θ)
Which well-known facts regarding sine and cosine are illustrated by the following?
θ
π −θ
θ
cos θ
sin θ
1
1
1
√
2
1
√
3
2
37
Solving Triangles A triangle is described by six values: three side lengths and three angle mea-
sures. Euclid’s triangle congruence theorems (SAS, ASA, SSS, SAA) say that three of these in suitable
combination are enough to recover the rest. In analytic geometry, these calculations typically use the
sine and cosine rules.
Theorem 3.7. Label the sides/angles of △ABC following the standard convention (page 37):
Sine Rule If d is the diameter of the circumcircle (Defn. 2.30), then
sin α
a
=
sin β
b
=
sin γ
c
=
1
d
Cosine Rule c
2
= a
2
+ b
2
−2ab cos γ
Proof. We prove the sine rule and leave the cosine rule as an exercise.
Everything relies on Corollary 2.32. Draw the circumcircle of △ABC.
Construct △BCD with diameter BD; this is right-angled at C by Thales’
Theorem. There are two cases:
1. If A and D lie on the same side of
←→
BC, then they share the same arc.
But then ∡BDC = α and
a = d sin ∡BDC = d sin α
2. If A and D lie on opposite sides of
←→
BC, then the quadrilateral ABDC
lies on a circle. Opposite angles at A, D are supplementary, whence
sin α = sin(π − α) = sin ∡BDC =
a
d
The two other angle-side combinations follow by permutation.
a
d
α
α
A
B
C
D
a
d
α
π −α
A
B
C
D
Examples 3.8. 1. Given SSS data, we may compute the three angles using the cosine rule. For
instance the given triangle has
α =
6
2
+ 7
2
−3
2
2 ·6 · 7
= cos
−1
19
21
≈ 25° β =
3
2
+ 7
2
−6
2
2 ·3 · 7
= cos
−1
11
21
≈ 58°
γ =
3
2
+ 6
2
−7
2
2 ·3 · 6
= cos
−1
−1
9
≈ 96°
Once you have α, you could alternatively switch to the sine rule to find β, before
computing γ = π − α − β.
7
3
6
α
β
γ
2. To solve the given triangle with ASA data, first find the remaining angle
γ = π −
π
4
−
π
3
=
5π
12
before applying the sine rule
sin
π
4
a
=
sin
π
3
b
= sin
5π
12
=⇒ a =
1
√
2 sin
5π
12
≈ 0.732
b =
√
3
2 sin
5π
12
≈ 0.897
1
a
b
π
4
π
3
γ
38
Multiple-angle formulæ The picture provides a very simple
proof of the expressions
sin(α + β) = sin α cos β + cos α sin β
cos(α + β) = cos α cos β − sin α sin β
at least when α + β <
π
2
. A little algebraic manipulation pro-
duces the double-angle and difference formulæ, and verifies
that these hold for all possible angle inputs.
α
β
1
α
α
cos β
sin β
sin α cos β
cos α sin β
sin 2α = 2 sin α cos α sin(α − β) = sin α cos β − cos α sin β
cos 2α = cos
2
α − sin
2
α = 2 cos
2
α − 1 cos(α − β) = cos α cos β + sin α sin β
Exercises 3.2. 1. A triangle has angle of
2π
3
radians between sides of lengths 2 and
√
3 −1. Find the
length of the remaining side, and the remaining angles.
2. Describe how to solve a triangle given data in line with the SAA congruence theorem.
3. Two measurements for the height of a mountain are taken at sea level 5000 ft apart in a line
pointing away from the mountain. The angles of elevation to the mountain top from the hori-
zontal are 15° and 13° respectively. What is the height of the mountain?
4. Use a multiple angle formula to find an exact value for cos
π
12
and thus exact values for the side
lengths of the triangle in Example 3.8.2.
5. The area of a triangle is
1
2
(base)·(height). By using each side of the triangle alternately as the
‘base,’ find an alternative proof of the sine rule without the relationship to the circumcircle.
6. You are given SSA data for a triangle: sides with lengths a = 1 and b =
√
3 and angle α =
π
6
.
Show that there are two triangles satisfying this data. Can you generalize to general SSA data?
7. (a) By dropping a perpendicular from B to
←→
AC at D and applying
the Pythagorean theorem, construct a proof of the cosine rule.
(b) Is your argument valid if D is not interior to AC? Explain.
a
c
b
h
γ
B
AC
D
8. The dot product of A = (a
1
, a
2
) and B = (b
1
, b
2
) is A · B := a
1
b
1
+ a
2
b
2
. Apply the cosine rule to
△OAB to prove that
A · B =
|
OA
||
OB
|
cos ∡AOB
9. Derive the multiple-angle formula for sin(α − β).
(Remember that 0 ≤ α, β, α − β < 2π so you can’t simply switch the sign of β!)
10. Given the arrangement pictured, find x, the radian-measure α and
the exact value of cos α.
(Hint: first show that you have similar isosceles triangles)
1
x
x x
1 − x
α
39
3.3 Isometries
At the heart of elementary geometry is congruence, the idea that geometric figures can be essentially
the same without necessarily being equal. In analytic geometry, congruence is described algebraically
using functions. This is motivated by the fact that congruent segments have the same length.
Definition 3.9. A function f : R
2
→ R
2
is a (Euclidean) isometry if it preserves lengths:
16
∀P, Q ∈ R
2
, d
f (P), f (Q)
=
|
PQ
|
Two figures (segments, angles, triangles, etc.) are said to be isometric (or congruent) precisely when
there is an isometry f : R
2
→ R
2
mapping one to the other.
Example 3.10. We check that the map f (x, y) =
1
5
3x + 4y, 4x − 3y
+ (3, 1) is an isometry. If
P = (x, y) and Q = (v, w), then
d
f (P), f (Q)
2
=
3v + 4w −3x −4y
5
2
+
4v −3w −4x + 3y
5
2
=
3
2
+ 4
2
5
2
(v − x)
2
+ (w − y)
2
=
|
PQ
|
2
Isometric segments are certainly congruent. We should make sure the same holds for angles.
Lemma 3.11. Isometries preserve (non-oriented) angles: if f : R
2
→ R
2
is an isometry, then
∠PQR
∼
=
∠ f (P) f (Q) f (R)
Proof. Since f is an isometry, the sides of △PQR and △f (P) f (Q) f (R) are mutually congruent in
pairs. The SSS triangle congruence theorem says that the angles are also mutually congruent.
This helps justify the term congruent in the Definition. Examples 3.13 and Exercise 8 expand this idea.
Example (3.10, cont). Warning: Isometries can reverse orientation!
In the picture,
∡ABC =
π
2
but ∡ f (A) f (B) f (C) =
3π
2
= 2π −∡ABC
A
B
C
f
Our next task is to confirm our intuition that isometries are rotations, reflections and translations.
Given an isometry f , define g(X) = f (X) − f (O), where O is the origin. Then g is an isometry
g(P) − g(Q) = f (P) − f (Q) =⇒ d
g(P) , g(Q)
= d
f (P), f (Q)
=
|
PQ
|
which moreover fixes the origin: g(O) = O. We conclude that every isometry f is the composition of
an origin-preserving isometry g followed by a translation “+C:”
f (X) = g(X) + C
16
In ancient Greek, iso-metros is literally same measure (length/distance).
40
It thus suffices to describe the origin-preserving isometries g. For these, we make two observations.
1. Suppose
|
OQ
|
= 1 and let X
r
= rQ for some r ∈ R. Then
• g(X
r
) is a distance
|
r
|
=
|
OX
r
|
from the origin O = g(O).
• g(X
r
) is a distance
|
1 −r
|
=
|
QX
r
|
from g(Q).
g( X
r
) therefore lies on two circles, which necessarily intersect at
a single point. We conclude that
g(rQ) = rg(Q)
The picture shows the case 0 < r < 1, where the uniqueness of
intersection follows from 1 =
|
r
|
+
|
1 −r
|
.
|r|
|1 −r|
|1 −r|
O
Q
X
r
g(Q)
g(X
r
)
2. g(1, 0) lies on the unit circle and therefore has the form
g(1, 0) = S
θ
:=
cos θ, sin θ
for some θ ∈ [0, 2π). By preservation of length and angle
(Lemma 3.11), any other point S
ϕ
= (cos ϕ, sin ϕ) on the unit
circle must be mapped to one of two points
g(S
ϕ
) = S
θ±ϕ
=
cos(θ ± ϕ), sin(θ ± ϕ)
The angle ϕ is therefore transferred to one side of the ray
−−→
OS
θ
.
0
1
y
0 1
x
S
ϕ
S
θ
S
θ+ϕ
S
θ−ϕ
ϕ
θ
By writing X = rS
ϕ
= (r cos ϕ, r sin ϕ) in polar co-ordinates and combining the above observations,
we conclude that g has one of two forms:
y
x
X = rS
ϕ
g(X) = rS
θ+ϕ
ϕ
θ
y
x
θ
2
ϕ
θ
2
−ϕ
θ
2
−ϕ
X = rS
ϕ
g(X) = rS
θ−ϕ
Rotation counter-clockwise by θ Reflection across the line making
angle
θ
2
with the positive x-axis
Theorem 3.12. Every isometry of R
2
has the form
f (X) = g(X) + C
where g is a rotation about the origin or a reflection across a line through the origin.
41
Calculating with isometries
This benefits from column-vector notation and matrix multiplication. Writing x =
(
x
y
)
=
r cos ϕ
r sin ϕ
for
the position vector of X
r
= (x, y) = rS
ϕ
and applying the multiple-angle formulæ, rotation becomes
g(x) = r
cos(θ + ϕ)
sin(θ + ϕ)
= r
cos θ cos ϕ −sin θ sin ϕ
sin θ cos ϕ + cos θ sin ϕ
=
cos θ −sin θ
sin θ cos θ
x
For reflections, the sign of the second column is reversed:
cos θ sin θ
sin θ −cos θ
. Every isometry therefore has
the form f (x) = Ax + c where A is an orthogonal matrix.
17
Examples 3.13. 1. We revisit Example 3.10 in matrix format:
f (x) =
1
5
3x + 4y
4x − 3y
+
3
1
=
1
5
3 4
4 −3
x
y
+
3
1
Since
sin θ
cos θ
=
4/5
3/5
=
4
3
, we see that its effect is to reflect across the line through the origin making
angle
1
2
tan
−1
4
3
≈ 26.6° with the positive x-axis, before translating by (3, 1).
2. △
a
has vertices (0, 0), (1, 0), (2, −1) and is congruent to △
b
, two of whose vertices are (1, 2) and
(1, 3). Find all isometries transforming △
a
to △
b
and the location(s) of the third vertex of △
b
.
Let f = Ax + c be the isometry. Since d
(1, 2), (1, 3)
= 1 these points must be the images
under f of (0, 0) and (1, 0). There are four distinct isometries:
Cases 1, 2: If f (0, 0) = (1, 2) and f (1, 0) = (1, 3), then c = f
0
0
=
1
2
and
A
1
0
+ c =
1
3
=⇒ A
1
0
=
0
1
=⇒ A =
0 a
12
1 a
22
for some a
12
, a
22
. Since A is orthogonal, the options are A =
0 ∓1
1 0
and
we obtain two possible isometries:
• f
1
(x) =
0 −1
1 0
x +
1
2
rotates by 90°, then translates by
1
2
.
• f
2
(x) =
0 1
1 0
x +
1
2
reflects across y = x, then translates by
1
2
.
The third point of △
b
is f
1
(2, −1) = ( 2, 4) or f
2
(2, −1) = ( 0, 4).
Cases 3, 4: f (0, 0) = (1, 3) and f (1, 0) = (1, 2) results in two further
isometries f
3
and f
4
. The details are an exercise.
All four possible triangles △
b
are drawn in the picture.
−1
0
1
2
3
4
1 2
△
a
△
b
In 1872, Felix Klein suggested that the geometry of a set is the study of its invariants: properties
preserved by its group of structure-preserving transformations. In Euclidean geometry, this is the
group of Euclidean isometries (Exercise 10). Klein’s approach provided a method for analyzing and
comparing the non-Euclidean geometries beginning to appear in the late 1800s. By the mid 1900s, the
resulting theory of Lie groups had largely classified classical geometries. Klein’s algebraic approach
remains dominant in modern mathematics and physics research.
17
An orthogonal matrix satisfies A
T
A = I. All such have the form
cos θ ∓sin θ
sin θ ±cos θ
=
a ∓b
b ±a
where a
2
+ b
2
= 1.
42
Exercises 3.3. 1. Let f : R
2
→ R
2
be the isometry, “reflect across the line through the origin making
angle
π
3
with the positive x-axis.” Find a 2 ×2 matrix A such that f (x) = Ax.
2. Describe the geometric effect of the isometry f (x) =
1
2
1
√
3
−
√
3 1
x +
3
−2
3. Find the remaining isometries f
3
, f
4
and the third points of △
b
in Exercise 3.13.2.
4. Find the reflection of the point ( 4, 1) across the line making angle
1
2
tan
−1
12
5
≈ 33.7° with the
positive x-axis.
(Hint: if tan θ =
12
5
, what are cos θ and sin θ?)
5. An origin-preserving isometry f (v) = Av moves the point (7, 4) to (−1, 8).
(a) If f is a rotation, find the matrix A. Through what angle does it rotate?
(b) If f is a reflection, find the matrix A. Across which line does it reflect?
6. Let ABCD be the rectangle with vertices A = (0, 0), B = (4, 0), C = (4, 3), D = ( 0, 3). Suppose
an isometry f : R
2
→ R
2
maps ABCD to a new rectangle PQRS where
P = f (A) := (2, 4) and R = f (C) := ( 2, 9)
Find all possible isometries f and the remaining points Q = f (B) and S = f (D).
7. (a) If A =
cos θ −sin θ
sin θ cos θ
and p is constant, explain why f (x) = A(x −p) + p = Ax + (I − A)p
rotates by θ around the point with position vector p.
(b) Suppose f (x) = Ax + c rotates the plane around the point P = (−2, 1) by an angle θ =
tan
−1
3
4
. Find A and c.
(c) Suppose f rotates by θ around p and g rotates by ϕ around q where θ, ϕ are non-zero.
i. If θ + ϕ = 2π, show that f ◦ g is a rotation: by what angle and about which point?
ii. What happens instead if θ + ϕ = 2π?
8. Suppose △ABC
∼
=
△PQR. Prove that there exists an isometry f : R
2
→ R
2
mapping one
triangle to the other. How many distinct such isometries could there be, and how does this
number depend on the triangles?
9. Make an argument involving circle intersections (see page 41) to prove that for any isometry f ,
f
(1 − t)P + tQ
= (1 −t) f (P) + t f (Q)
10. Throughout this question, we use the notation f
A,c
: x 7→ Ax + c.
(a) Prove that isometries obey the composition law f
A,c
◦ f
B,d
= f
AB,c+Ad
.
(b) Find the inverse function of the isometry f
A,c
. Otherwise said, if f
A,c
◦ f
C,d
= f
I,0
, where I
is the identity matrix, how do B, d depend on A, c?
(c) Verify that the following composition f
A,c
◦ f
I,d
◦ f
−1
A,c
is a translation.
Part (a) can be written using augmented matrices: (A |c)(B |d) := (AB |c + Ad).
If you know group theory, parts (a) and (b) are the closure and inverse properties for the group of Eu-
clidean isometries E. Part (c) says that the translations T form a normal subgroup; E is therefore a
semi-direct product of T and the orthogonal group of origin-preserving isometries E = T ⋊ O
2
(R ).
43
3.4 The Complex Plane
Complex numbers date to 16
th
century Italy. Their application to geometry really begins with Leon-
hard Euler (1707–1783) who identified the set of complex numbers C with the plane (what is now
known as the Argand diagram).
Definition 3.14. Let i be an abstract symbol satisfying the property
i
2
= −1.
Given real numbers x, y, the complex number z = x + iy is simply the
point (x, y) in the standard Cartesian plane.
18
Given z = x + iy, its:
• Complex conjugate z = x −iy is its reflection across the real axis.
• Modulus
|
z
|
=
√
zz =
p
x
2
+ y
2
is its distance from the origin.
• Argument arg(z) is the angle (measured counter-clockwise) be-
tween the positive real axis and the ray
−→
0z.
y
1 2 3 4
z = 2 −3i
z = 2 + 3i
|z| =
√
13
x
i
2i
3i
−i
−2i
−3i
arg(z) = tan
−1
3
2
Addition, scalar multiplication (by real numbers) and complex multiplication follow the usual algebraic
rules while using i
2
= −1 to simplify.
Example 3.15. A simple example of multiplication of complex numbers:
(2 + 3i)(4 + 5i) = 2 ·4 + 2 ·5i + 3i ·4 + 3i ·5i (multiply out)
= 8 + 10i + 12i −15 (use i
2
= −1 to simplify)
= −7 + 22i
The algebra screams geometry! Definition 3.14 already length, angle and reflection in the real axis.
Two other aspects of basic geometry are immediate:
• Addition by z translates all points by z.
• Scalar multiplication scales distances from the origin (similarity).
The algebraic property distinguishing the complex numbers from the standard Cartesian plane is
complex multiplication. To start visualizing this, consider multiplication by i,
iz = i(x + iy) = −y + ix
This is the result of rotating z counter-clockwise
π
2
radians about the origin. To obtain all rotations
and reflections, we need an alternative description of a complex number.
Lemma 3.16. 1. (Euler’s Formula) For any θ ∈ R, e
iθ
= cos θ + i sin θ.
2. (Exponential laws) e
iθ
e
iϕ
= e
i(θ+ϕ)
and (e
iθ
)
n
= e
inθ
for any n ∈ Z.
Evaluating at θ = π yields the famous Euler identity e
iπ
= −1. Part 1 can be taken as a definition. To
see that it is a reasonable definition requires either power series or elementary differential equations,
topics best described elsewhere. Part 2 is an exercise.
18
In the language of linear algebra, C is a vector space over R with basis {1, i}.
44
Definition 3.17. Let z = x + iy be a non-zero complex number.
Writing x = r cos θ and y = r sin θ, we obtain the polar form
z = re
iθ
= r(cos θ + i sin θ)
where r =
|
z
|
is the modulus and θ = arg(z) the argument of z.
0 1 2
1 +
√
3i = 2e
iπ
3
2
π
3
i
0
2i
Now consider the effect of multiplying a complex number z = re
iϕ
by e
iθ
= cos θ + i sin θ: according
to the Lemma
e
iθ
z = re
iθ
e
iϕ
= re
i(θ+ϕ)
which has the same modulus (r) as z but a new argument.
Theorem 3.18. The complex number e
iθ
z is the result of rotating z counter-clockwise about the origin
through an angle θ.
Example 3.19. To rotate z = 1 + 2i counter-clockwise by
3π
4
radians, we multiply by
e
3πi
4
= cos
3π
4
+ i sin
3π
4
=
1
√
2
(−1 + i)
to obtain
e
3πi
4
z =
1
√
2
(−1 + i)(1 + 2i) = −
1
√
2
(3 + i)
−2 −1 1
z
e
3πi
4
z
3π
4
i
2i
−i
You could try to keep things in polar form, though it doesn’t result in a nice answer:
z =
√
5e
i tan
−1
2
=⇒ e
3πi
4
z =
√
5e
3πi
4
+i tan
−1
2
Reflections may be described by combining rotations with complex con-
jugation. To reflect across the line making angle θ with the positive real
axis, we rotate the plane so that the reflection appears to be vertical:
1. Rotate the plane clockwise by θ, that is z 7→ e
−iθ
z.
2. Reflect across the real axis by complex conjugation.
3. Rotate counter-clockwise by θ.
Combining these steps gives the formula.
ϕ
1
2
3
Theorem 3.20. To reflect z across the line making angle θ with the positive real axis, we compute
z 7→ e
iθ
( e
−iθ
z) = e
2iθ
z
45
Example 3.21. Reflect z = −2 + 3i across the line through the origin and w =
√
3 + i.
First compute θ = arg(w) = tan
−1
1
√
3
=
π
6
. The desired point is therefore
e
iπ
3
(−2 −3i) =
1
2
+
√
3
2
i
!
(−2 −3i) =
3
√
3
2
−1
!
−
√
3 +
3
2
i
To describe general rotations and reflections about arbitrary points/lines, we combine our approach
with translations (compare Exercise 3.3.7).
Corollary 3.22. 1. To rotate z by θ about a point w, compute z 7→ e
iθ
(z −w) + w.
2. To reflect z across the line with slope θ through a point w, compute z 7→ e
2iθ
( z −w) + w.
Example 3.23. The combination of translation by −i, rotation by
π
3
around the origin, then translation
by 1, may be expressed
z 7→ e
π
3
z − i
+ 1 = i + e
π
3
z − i
+ 1 − i
Alternatively, this is rotation by
π
3
around i followed by translation by 1 − i.
We have now described all the Euclidean isometries of the previous section in the language of com-
plex numbers. Here is the full dictionary.
19
Isometry/Transformation Complex numbers Matrices/vectors
Addition/Translation z + w = (x + iy) + (u + iv) z + w =
x
y
+
u
v
Scaling λz = (λx) + i(λy) λz =
λx
λy
Rotation CCW by
π
2
z 7→ iz z 7→
0 −1
1 0
z
Rotation CCW by θ z 7→ e
iθ
z z 7→
cos θ −sin θ
sin θ cos θ
z
Vertical reflection z 7→ z z 7→
1 0
0 −1
z
Reflection across line with slope
θ
2
z 7→ e
iθ
z z 7→
cos θ sin θ
sin θ −cos θ
z
It is perhaps surprising to modern readers, but complex numbers came before vectors and matrix-
geometry! During the 1800s mathematicians tried unsuccessfully to replicate the complex number
approach in higher dimensions. This ultimately led (via Hamilton’s quaternions) to the adoption of
vectors and linear algebra/matrix calculations.
One reason for the desire to keep the complex number description is that it may be used to describe
further (non-isometric) transformations of the plane: for instance z 7→ z
−1
is reflection in a circle! We’ll
discuss some of this at the end of Chapter 4.
19
Scaling isn’t an isometry, but it is worth including nonetheless!
46
Exercises 3.4. 1. Use complex numbers to compute the result of the following transformations: you
can answer in either standard or polar form.
(a) Rotate 3 −5i counter-clockwise around the origin by
3π
4
radians.
(b) Reflect 2 −i across the line joining 1 + i
√
3 and the origin.
(c) Reflect 1 + i across the line through the origin making angle
π
5
radians with the positive
real axis.
2. Find the reflection of the point (2, 3) across the line through the origin making angle
3π
8
with
the positive x-axis. Give your answer using both complex numbers and matrices/vectors.
3. Repeat the previous question for the point (3, 4) and the angle
5π
12
= 75°.
4. Describe the geometric effect of the map z 7→
1
√
2
(−1 −i)
z − 3 + 4i
.
(Hint: compare Example 3.23)
5. (Hard) Consider the line ℓ through the origin and
p
2 +
√
2,
p
2 −
√
2
. Compute the result
of reflecting −2 + 3i across ℓ.
6. By letting n = 3 in Lemma 3.16, prove that
cos 3θ = 4 cos
3
θ − 3 cos θ
Find a corresponding trigonometric identity for sin 3θ.
7. Prove part 2 of Lemma 3.16.
(Hint: use the multiple-angle formulae (page 39) to expand e
i(θ+ϕ)
)
47
3.5 Birkhoff’s Axiomatic System for Analytic Geometry (non-examinable)
Recall that analytic geometry, as developed by Descartes and Fermat, was originally conceived as a
bolt-on to Euclidean geometry. In 1932 George David Birkhoff provided an axiomatization of analytic
geometry in its own right.
Background Assume the usual properties/axioms of the real numbers as a complete ordered field.
Birkhoff’s approach is typical of modern axiomatic systems in that it is built on top of pre-existing
systems (set theory, complete ordered fields, etc.).
Undefined terms Two objects: Point, line. Two functions: distance d, angle measure ∡. If the set of
points is S, then,
d : S ×S → R
+
0
, ∡ : S × S × S → [0, 2π)
Axioms Euclidean Given two distinct points, there exists a unique line containing them.
Ruler Points on a line ℓ are in bijective correspondence with the real numbers in such a way that if
t
A
, t
B
correspond to A, B ∈ ℓ, then
|
t
A
−t
B
|
= d(A, B).
Protractor The rays emanating from a point O are in bijective correspondence with the set [0, 2π) so
that if α, β correspond to rays
−→
OA,
−→
OB, then ∡AOB ≡ β −α (mod 2π). This correspondence is
continuous in A, B.
SAS similarity
20
If △ABC and △XYZ satisfy ∡ABC = ∡XYZ and
d(A,B)
d(X,Y)
=
d(B,C)
d(Y,Z)
, then the remaining
angles have equal measure and the final sides are in the same ratio (i.e., △ABC ∼ △XYZ).
Definitions As with Hilbert, some of these are required before later axioms make sense. In partic-
ular, the definition of ray is required before the protractor axiom.
Betweenness B lies between A and C if d(A, B) + d(B, C) = d(A, C)
Segment AB consists of the points A, B and all those between
Ray
−→
AB consists of the segment AB and all points C such that B lies between A and C.
Basic shapes Triangles, circles, etc.
Analytic Geometry as a Model
The axioms should feel familiar. Being shorter than Hilbert’s list, and being built on familiar notions
such as the real line, it is somewhat easier for us to understand what the axioms are saying and to
visualize them. There is something to prove however; indeed the major point of Birkhoff’s system!
Theorem 3.24. Cartesian analytic geometry is a model of Birkhoff’s axioms.
Recall what this requires: we must provide a definition of each of the undefined terms and prove that
these satisfy each of Birkhoff’s axioms. Here are suitable definitions for Cartesian analytic geometry:
20
As with Hilbert, Birkhoff makes SAS an axiom: Birkhoff’s version is stronger in that it also applies to similar triangles.
48
Point An ordered pair (x, y) of real numbers.
Distance d(A, B) =
q
(A
x
− B
x
)
2
+ (A
y
− B
y
)
2
Line All points satisfying a linear equation ax + by + c = 0.
Angle Define column vectors as differences (v = P − O and w = Q − O) and consider the matrix
J =
0 −1
1 0
. Now define angle via
cos ∡POQ =
v · w
|
v
||
w
|
where ∡POQ ∈
(
[0, π] ⇐⇒ w · Jv ≥ 0
(π, 2π) ⇐⇒ w · Jv < 0
(∗)
In essence J is ‘rotate counter-clockwise by
π
2
.’ Cosine may be defined using power series, so
no pre-existing geometric meaning is necessary.
Proof. (Euclidean axiom) If (x
1
, y
1
) and (x
2
, y
2
) satisfy ax + by + c = 0 then
a(x
1
− x
2
) + b(y
1
−y
2
) = 0
whence a = y
1
−y
2
, b = x
2
− x
1
up to scaling. It follows that the line has equation
(y
1
−y
2
)x + (x
2
− x
1
)y + x
1
y
2
− x
2
y
1
= 0
unique up to multiplication of all three of a, b, c by a non-zero constant.
The remaining axioms are exercises.
Exercises 3.5. 1. Prove that the ruler axiom is satisfied:
(a) First show that if P = Q lie on ℓ, then any point A on the line has the form
A = P +
t
A
d(P, Q)
(Q − P) where t
A
∈ R
(b) Use this formula to verify that d(A, B)
2
= (t
A
−t
B
)
2
.
2. Let i =
1
0
. Given any non-zero point B, define b = B −O and let β = cos
−1
i·b
|
b
|
in accordance
with (∗). This is a continuous function of b.
(a) If
ˆ
B is any other point on the same ray
−→
OB, explain why we get the same value β.
(β is thus a continuous function of B)
(b) If B = (x, y), what are values cos β and sin β?
(c) Suppose A corresponds to α under this identification. Evaluate cos(β − α) and therefore
prove that the protractor axiom is satisfied.
3. Use the cosine rule (Theorem 3.7) to prove that the SAS similarity axiom is satisfied.
49
