<<

. 2
( 13)



>>

between 0 and 2π. The product of
two complex numbers z and t is the
geometric operation consisting in the
rotation of the parallelogram
representing the first complex
Figure 3.3
number until it touches the
parallelogram representing the second
complex number. When they contact
in a unitary vector v (figure 3.3), the parallelogram formed by the other two vectors is the
product of both complex numbers:

v2 = 1
z=uv t=vw

zt=uvvw=uw

This geometric construction is always possible because a parallelogram can be
lengthened or widened maintaining the area so that one side has unity length.
The conjugate of a complex number (symbolised with an asterisk) is that number
whose imaginary part has opposite sign:

z* = a ’ b e12
z = a + b e12

The geometric meaning of the
conjugation is a permutation of the
vectors whose product is the complex
number (figure 3.4). In this case, the
inner product is preserved while the
outer product changes the sign. The
product of a complex number and its
Figure 3.4
conjugate is the square of the modulus:

z z* = u v v u = u2 = ¦z¦2

The quotient of complex numbers is defined as the product by the inverse. The
inverse of a complex number is equal to the conjugate divided by the square of the
modulus:
z * a ’ b e12
z ’1 = =2
a + b2
2
z

Then, let us see an example of quotient of complex numbers:

’ 2 + 5 e12 (’ 2 + 5 e12 )(3 + 4 e 12 ) ’ 26 + 7 e12
= =
3 + (’ 4 )
3 ’ 4 e12 2
25
2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 17


With the polar form, the quotient is obtained by dividing moduli and subtracting
arguments:

«z
z
=¬ ·
±
¬t·
t  ± ’ β
β



The best way to calculate the power of a complex number with natural exponent is
through the polar form, although for low exponents the binomial form and the Newton
formula is often used, e.g.:

( 2 ’ 3 e12 )3 = 23 ’ 3 · 22 e12 + 3 · 2 e122 ’ e123 = 8 ’ 12 e12 ’ 6 + e12 = 2 ’ 11 e12

From the characteristic property of the exponential function it follows that the
modulus of a power of a complex number is the power of its modulus, and the argument
of this power is the argument of the complex number multiplied by the exponent:

( )
n n
=z
z n±
±


This is a very useful rule for large exponents, e.g.:

( 2 + 2 e12 )1000 = ( 2 2 π/4 )1000 = [( 2 2 )1000]250 π = 215000 = 21500

When the argument exceeds 2π, divide by this value and take the remainder, in order to
have the argument within the period 0<± <2π.
Since a root is the inverse operation of a power, its value is obtained by extracting
the root of the modulus and dividing the argument by the index n. But a complex number
of argument ± may be also represented by the arguments ± +2 π k. Their division by the
index n yields n different arguments within a period, corresponding to n different roots:

z± = n z k = 0, ... n ’1
n
(± + 2π k ) / n



8 0 = { 2 0 , 2 2π / 3 , 2 4π/3 } . In the complex plane, the
For example, the cubic roots of 8 are 3

n-th roots of every complex number are located at the n vertices of a regular polygon.


Permutation of complex numbers and vectors

The permutative property of the vectors is intimately related with the commutative
property of the product of complex numbers. Let z and t be complex numbers and a, b, c
and d vectors fulfilling:

z=ab t=cd

Then the following equalities are equivalent:


zt=tz abcd=cdab
18 RAMON GONZALEZ CALVET


A complex number z and a vector c do not commute, but they can be permuted by
conjugating the complex number:

z c = a b c = c b a = c z*

Every real number commute with any vector. However every imaginary number
anticommute with any vector, because the imaginary unity e12 anticommute with e1 as
well as with e2:

zc=’cz z imaginary


The complex plane

In the complex plane, the complex numbers are represented taking the real
component as the abscissa and the imaginary component as the ordinate. The vectorial
plane differs from the complex plane in the fact that the vectorial plane is a plane of
absolute directions whereas the complex plane is a plane of relative directions with
respect to the real axis, to which we may assign any direction. As explained in more detail
in the following chapter, the unitary complex numbers are rotation operators applied to
vectors. The following equality shows the ambivalence of the Cartesian coordinates in the
Euclidean plane:

e1 ( x + y e12 ) = x e1 + y e 2

Due to a careless use, often the complex numbers have been improperly thought as
vectors on the plane, furnishing the confusion between the complex and vector planes to
our pupils. It will be argued that this has been very fruitful, but this argument cannot
satisfy geometers, who search the fundamentals of the geometry. On the other hand, some
physical magnitudes of a clearly vectorial kind have been taken improperly as complex
numbers, specially in quantum mechanics. Because of this, I™m astonished when seeing
how the inner and outer products transform in a special commutative and anti-
commutative products of complex numbers. The relation between vectors and complex
numbers is stated in the following way: If u is a fixed unitary vector, then every vector a
is mapped to a unique complex z fulfilling:

u2 = 1
a=uz with

Also other vector b is mapped to a complex number t:

b=ut

The outer and inner products of the vectors a and b can be written now using the
complex numbers z and t:

1 1 1
( a b ’ b a ) = ( u z u t ’ u t u z ) = u2 ( z* t ’ t* z ) =
a§b=
2 2 2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 19


1
( z* t ’ t* z ) = ( zR tI ’ zI tR ) e12
=
2

1 1 1
a·b= ( a b + b a ) = ( u z u t + u z u t ) = ( z* t + t* z ) = zR tR + zI tI
2 2 2

where zR, tR, zI, tI are the real and imaginary components of z and t. These products have
been called improperly scalar and exterior products of complexes. So, I repeat again that
complex quantities must be distinguished from vectorial quantities, and relative directions
(complex numbers) from absolute directions (vectors). A guide for doing this is the
reversion, under which the vectors are reversed while the complex number are not1.


Complex analytic functions

The complex numbers are a commutative algebra where we can study functions as
for the real numbers. A function f(x) is said to be analytical if its complex derivative
exists:
lim f (z ) ’ f (z 0 )
f(z) analytical at z0 ” ∃
z ’ z0 z ’ z0

This means that the derivative measured in any direction must give the same result. If f(x)
= a + b e12 and z = x + y e12 , the derivatives following the abscissa and ordinate directions
must be equal:

‚a ‚b ‚a ‚b
f' (z ) = + e12 = ’ e12 +
‚x ‚x ‚y ‚y

whence the Cauchy-Riemann conditions are obtained:

‚a ‚b ‚a ‚b
= =’
and
‚x ‚y ‚y ‚x

Due to its linearity, now the derivative in any direction are also equal. A consequence of
these conditions is the fact that the sum of both second derivatives (the Laplacian)
vanishes, that is, both components are harmonic functions:

‚ 2a ‚ 2a ‚ 2b ‚ 2b
+ = + =0
‚x 2 ‚y 2 ‚x 2 ‚y 2


1
A physical example is the alternating current. The voltage V and intensity I in an electric circuit
are continuously rotating vectors. The energy E dissipated by the circuit is the inner product of
both vectors, E = V · I. The impedance Z of the circuit is of course a complex number (it is
invariant under a reversion). The intensity vector can be calculated as the geometric product of the
voltage vector multiplied by the inverse of the impedance I = V Z’1 . If we take as reference a
continuously rotating direction, then V and I are replaced by pseudo complex numbers, but
properly they are vectors.
20 RAMON GONZALEZ CALVET


The values of a harmonic function (therefore the value of f(z)) within a region are
determined by those values at the boundary of this region. We will return to this matter
later. The typical example of analytic function is the complex exponential:

exp( x + y e12 ) = exp( x )(cos y + e12 sin y )

which is analytic in all the plane. The logarithm function is defined as the inverse function
of the exponential. Since z = ¦z¦exp(e12 •) where • is the argument of the complex, the
principal branch of the logarithm is defined as:

0 ¤ • < 2π
log z = log z + e12•

Also • + 2πk (k integer) are valid arguments for z yielding another branches of the
logarithm2. In Cartesian coordinates:

x
log( x + y e12 ) = log x 2 + y 2 + e12 arccos 0 ¤ • <π
x +y
2 2


« 
x
log( x + y e12 ) = log x + y + e12 ¬ arccos +π · π ¤ • < 2π
2 2
¬ ·
x +y
2 2
 

At the positive real half axis, this logarithm is not analytic because it is not continuous.
Now let us see the Cauchy™s theorem: if a function is analytic in a simply
connected domain on the complex plane, then its integral following a closed way C
within this domain is zero. If the analytic function is f (z ) = a + b e12 then the integral is:


« f (z ) dz = « (a + b e ) (dx + dy e ) = « (a dx ’ b dy ) + e « (a dy + b dx )
12 12 12
C C C C


Since C is a closed way, we may apply the Green theorem to write:

« ‚b ‚a  « ‚a ‚b 
= ’ «« ¬ +
¬ ‚x ‚y · dx dy + e12 «« ¬ ‚x ’ ‚y · dx dy = 0
· ¬ ·
D  D 

where D is the region bounded by the closed way C. Since f(z) fulfils the analyticity
conditions everywhere within D, the integral vanishes.
From here the following theorem is deduced: if f(z) is an analytic function in a
simply connected domain D and z1 and z2 are two points of D, then the definite integral
between these points has a unique value independently of the integration trajectory,
which is equal to the difference of the values of the primitive F(z) at both points:

dF (z )
f (z ) =
f (z ) dz = F (z 2 ) ’ F (z1 )
z2
« if
dz
z1




2 n
z.
The logarithm is said to be multi-valued. This is also the case of the roots
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 21


If f(z) is an analytic function (with a unique value) inside the region D bounded by
the closed path C, then the Cauchy integral formula is fulfilled for a counterclockwise
path orientation:

f (z )
1
dz = f (z 0 )
« z ’ z0
2π e12 C


Obviously, the integral does not vanish because the integrand is not analytic at z0 .
However, it is analytic at the other points of the region D, so that the integration path from
its beginning to its end passes always through an analyticity region, and by the former
theorem the definite integral must have a constant value, independently of the fact that
both extremes coincide. Now we integrate following a circular path z = z 0 + r exp(e12• ) ,
where the radius r is a real constant and the angle • is a real variable. The evaluation of
the integral gives f(z0):


1
« f (z + r exp(e12• )) d• = f (z 0 )
0
2π 0



because we can take any radius and also the limit r ’ 0. The consequence of this theorem
is immediate: the values of f(z) at a closed path C determine its value at any z0 inside the
region bounded by C. This is a characteristic property of the harmonic functions, already
commented above.
Let us rewrite the Cauchy integral formula in a more suitable form:

f (t )
1
« t ’ z dt = f (z )
2π e12 C


The first and successive derivative with respect to z are:

f (t ) f (t )
n!
1
dt = f' (z ) (z )
« (t ’ z ) « (t ’ z ) dt = f n
n +1
2
2π e 12 2π e12
C C


For z = 0 we have:

f (t ) f (t )
n!
1
dt = f (0) (0)
« « dt = f n
n +1
2π e12 t 2π e12 t
C C


Now we see that these integrals always exist if f(z) is analytic, that is, all the derivatives
exist at the points where the function is analytic. In other words, the existence of the first
derivative (analyticity) implies the existence of those with higher order.
The Cauchy integral formula may be converted into a power series of z:

f (t ) dt
f (t ) f (t ) ∞ « z 
k
1 1 1
f (z ) = ‘ ¬ · dt
« t ’ z 2π e12 « t (1 ’ z / t ) 2 π e12 «
dt = =
2 π e12 t k =0  t 
C C C


Rewriting this expression we find the Taylor series:
22 RAMON GONZALEZ CALVET



f (t ) f k (0) k
∞ ∞
1
f (z ) = ‘z « dt = ‘
k
z
t k +1
2 π e12 k!
k =0 k =0
C


The only assumption made in the deduction is the analyticity of f(z). So this series is
convergent within the largest circle centred at the origin where the function is analytic
(that is, the convergence circle touches the closest singular point). The Taylor series is
unique for any analytic function. On the other hand, every analytic function has a Taylor
series.
Instead of the origin we can take a series centred at another point z0. In this case,
following the same way as above, one arrives to the MacLaurin series:

(z 0 )
k

f
f (z ) = ‘ ( z ’ z 0 )k
k!
k =0



For instance, the Taylor series (z0 = 0) of the exponential, taking into account the fact that
all the derivatives are equal to the exponential and exp(0) = 1, is:

z2 z3
exp(z ) = 1 + z + + + ...
2 ! 3!

The exponential has not any singular point. Then the radius of convergence is infinite.
In order to find a convergent series for a function which is analytic in an annulus
although not at its centre (for r1 <¦z ’ z0¦< r2 as shown in figure 3.5), we must add
powers with negative exponents, obtaining the Lauren series:


f (z ) = ‘ a (z ’ z ) Figure 3.5
k
k 0
k = ’∞


The Lauren series is unique, and coincident
with the McLaurin series if the function has
not any singularity at the central region. The
coefficients are obtained in the same way as
above:

f (t ) dt
1
«
ak =
t k +1
2 π e12 C


where the path C encloses the central circle. When f(x) is not analytic at some point of this
central circle (e.g. z0), the powers of negative exponent appear in the Laurent series. The
coefficient a’1 is called the residue of f. The Lauren series is the addition of a series of
powers with negative exponent and the McLaurin series:

∞ ’1 ∞
f (z ) = ‘ a (z ’ z ) ‘ a (z ’ z ) + ‘ a (z ’ z )
k k k
=
k 0 k 0 k 0
k = ’∞ k = ’∞ k =0
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 23


and is convergent only when both series are convergent, so that the annulus goes from the
radius of convergence of the first series (r1) to the radius of convergence of the McLaurin
series (r2).
Let us review singular points, the points where an analytic function is not defined.
An isolated singular point may be a removable singularity, an essential singularity or a
pole. A point z0 is a removable singularity if the limit of the function at this point exists
and, therefore, we may remove the singularity taking the limit as the value of the function
at z0. A point z0 is a pole of a function f(z) if it is a zero of the function 1/f(z). Finally, z0 is
an essential singularity if both limits of f(z) and 1/f(z) at z0 do not exist.
The Lauren series centred at a removable singularity has not powers with negative
exponent. The series centred at a pole has a finite number of powers with negative
exponent, and that centred at an essential singularity has an infinite number of powers
with negative exponent. Let us see some examples. The function sin z / z has a removable
singularity at z = 0:

z2 z4 z6
sin z
=1’ + ’ + ...
z 3! 5 ! 7 !

So the series has only positive exponents.
The function exp(1/z) has an essential singularity at z = 0 . From the series of
exp(z), we obtain a Lauren series with an infinite number of powers with negative
exponent by changing z for 1/z. Also the radius of convergence is infinite:

«1 1 1 1
exp¬ · = 1 + + + + ...
z 2 ! z 2 3! z 3
z

Finally, the function 1/ z2 (z ’1) has a pole at z = 0 . Its Lauren series:

= ’ 2 (1 + z + z 2 + z 3 + z 4 ...) = ’ 2 ’ ’ 1 ’ z ’ z 2 ...
1 1 11
z 2 (z ’ 1) z
z z

is convergent for 0<¦z¦<1 since the function has another pole at z = 1.
To see the importance of the residue, let us calculate the integral of a function
through an annular way from its Lauren series:



« f (z ) dz = ‘ a « (z ’ z )
k
dz
k 0
k = ’∞
C C



For k≥0 the integral is zero because (z ’ z0)k is analytic in the whole domain enclosed by
C. For k<’2 the integral is also zero because the path is inside a region where the powers
are analytic:

lim ® (z ’ z 0 )k ’1 
z2


« (z ’ z )
k
dz =  =0

0
z1 ’ z 2 ° k ’ 1 »
C z1
24 RAMON GONZALEZ CALVET


However k = ’1 is a special case. Taking the circular path z = r exp(e12• ) ( r1 < r < r2 )
we have:

dz
« z ’ z 0 = e12 « d• = 2π e12
C 0


So that the residue theorem is obtained:

« f (z ) dz = 2π e a ’1
12
C



where a’1 is the coefficient of the Lauren series centred at the pole. If the path C encloses
some poles, then the integral is proportional to the sum of the residues:

« f (z ) dz = 2π e ‘ residues
12
C


Let us see the case of the last example. If C is a path enclosing z = 0 and z = 1 then the
residue for the first pole is 1 and that for the second pole “1 (for a counterclockwise path)
so that the integral vanishes:

«1 1 1
dz dz dz
« = «¬ ’ ’ 2 · dz = « ’« =0
z 2 (z ’ 1) C  z ’ 1 z z  z ’1 C z
C C




The fundamental theorem of algebra

Firstly let us prove the Liouville™s theorem: if f(x) is analytic and bounded in the
whole complex plane then it is a constant. If f(x) is bounded we have:

¦f(x)¦< M

The derivative of f(x) is always given by:

f (t )
1
f' (z ) = « (t ’ z ) dt
2
2 π e 12 C



Following the circular path t ’ z = r exp(e12• ) we have:


1
f' (z ) = « f (r exp(e • )) exp(’ e • ) d•
12 12
2π r 0




« f ( z ) dz ¤ «
Using the inequality f ( z ) dz , we find:

2π 2π
1 1 M
f' (z ) ¤ f (r exp(e12• ))
« « M d• =
d• ¤
2π r 2π r 2π r
0 0
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 25



Since the function is analytic in the entire plane, we may take the radius r as large as we
wish. In consequence, the derivative must be null and the function constant, which is the
proof of the theorem.
A main consequence of the Liouville™s theorem is the fundamental theorem of
algebra: any polynomial of degree n has always n zeros (not necessarily different):

p(z) = a0 + a1 z + a2 z2 + ... + an zn = 0 ’ ∃ zi i ∈{1, .., n} p(zi) = 0

where a0, a1, a2, ... an are complex coefficients. For real coefficients, the zeros are whether
real or pairs of conjugate complex numbers. The proof is by supposing that p(z) has not
any zero. In this case f(z) = 1/p(z) is analytic and bounded (because p(z)’ 0 for ¦z¦’ ∞)
in the whole plane. From the Liouville™s theorem f(z) and p(z) should be constant
becoming in contradiction with the fact that p(z) is a polynomial. In conclusion p(z) has at
least one zero.
According to the division algorithm, the division of the polynomial p(z) by z ’ b
decreases the degree of the quotient q(z) by a unity, and yields a complex number r as
remainder:

p(z) = (z ’ b) q(z) + r

The substitution of z by b gives:

p(b) = r

That is, the remainder of the division of a polynomial by z ’ b is equal to its numerical
value for z = b . On the other hand, if b is a zero zi, the remainder vanishes and we have an
exact division:

p(z) = (z ’ zi) q(z)

Again q(z) has at least one zero. In each division, we find a new zero and a new factor, so
that the polynomial completely factorises with as many zeros and factors as the degree of
the polynomial, which ends the proof:

n
p( z ) = a n ∏ ( z ’ z i ) n = degree of p(z)
i =1



For example, let us calculate the zeros of the polynomial z3 ’ 5 z2 + 8 z ’ 6. By the
Ruffini method we find the zero z = 3:

¦1 ’5 ’6
8
¦
¦ ’6
3 3 6
°_______________________
’2 2¦0
1

The zeros of the quotient polynomial z2 ’2z ’2 are obtained through the formula
of the equation of second degree:
26 RAMON GONZALEZ CALVET



2 ± 4 ’ 4 …1… 2
’ 1 = 1 ± e12
z= =1±
2 …1

yielding a pair of conjugate zeros. Then the factorisation of the polynomial is:

z3 ’ 5 z2 + 8 z ’ 6 = ( z ’ 3 ) ( z ’ [1 + e12 ] ) ( z ’ [1 ’ e12 ] )


Exercises

3.1 Multiply the complex numbers z = 1 + 3 e12 and t = ’2 + 2 e12. Draw the geometric
figure of their product and check the result found.

3.2 Prove that the modulus of a complex number is the square root of the product of this
number by its conjugate.

3.3 Solve the equation x4 ’1 = 0. Being of fourth degree, you must obtain four complex
solutions.

3.4 For which natural values of n is fulfilled the following equation?

( 1 + e12 )n + ( 1 ’ e12 )n = 0

3.5 Find the cubic roots of ’3 + 3 e12 .

3.6 Solve the equation:

z2 + ( ’3 + 2 e12 ) z + 5 ’ e12 = 0

3.7 Find the analytical extension of the real functions sin x and cos x.

3.8 Find the Taylor series of log(1 + exp(’ z )) .
1
in the annulus 1<¦z ’ 2 ¦< 4.
3.9 Calculate the Lauren series of 2
z + 2z ’ 8

1
3.10 Find the radius of convergence of the series ‘ n and its analytic function.
4 (z + 1)
n
n =1

sin z
3.11Calculate the Lauren series of 2 and the annulus of convergence.
z

3.12 Prove that if f(z) is analytic and does not vanish then it is a conformal mapping.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 27


4. TRANSFORMATIONS OF VECTORS

The transformations of vectors are mappings from the vector plane to itself. Those
transformations preserving the modulus of vectors, such as rotations and reflections, are
called isometries and those which preserve angles between vectors are said to be
conformal. Besides rotations and reflections, the inversions and dilatations are also
conformal transformations.
Figure 4.1
Rotations

A rotation through an angle ± is the
geometric operation consisting in turning a
vector until it forms an angle ± with the
previous orientation. The positive direction of
angles is counterclockwise (figure 4.1). Under
rotations the modulus of any vector is
preserved. According to the definition of
geometric product, the multiplication of a
vector v by a unitary complex number with
argument ± produces a vector v' rotated
through an angle ± with regard to v.

v' = v 1± = v ( cos ± + e12 sin ± )

This algebraic expression for rotations, when applied to a real or complex number
instead of vector, modifies its value. However, real numbers are invariant under rotations
and the parallelograms can be turned without changing the complex number which they
represent. Therefore this expression for rotations, although being useful for vectors, is not
valid for complex numbers. In order to remodel it, we factorise the unitary complex
number into a product of two complex numbers with half argument. According to the
permutative property, we can permute the vector and the first complex number whenever
writing the conjugate:

± ± ± ±
’ e12 sin
v' = v 1± = v 1±/2 1±/2 = 1’±/2 v 1±/2 = ( cos ) v ( cos + e12 sin )
2 2 2 2

The algebraic expression for rotations now found preserves complex numbers:

z' = 1’±/2 z 1±/2 = z

Let us calculate the rotation of the vector 4 e1 through 2π / 3 by multiplying it by the
unitary complex with this argument:

«1 3
« 2π 

= 4 e1 ¬ ’ + e12 · = ’2 e1 + 2 3 e 2
v' = 4 e1 ¬ cos + e12 sin · ¬2 2·
3 3
  

On the other hand, using the half angle π/3 we have:
28 RAMON GONZÁLEZ CALVET


« π « π
π π
v' = ¬ cos ’ e12 sin · 4 e1 ¬ cos + e12 sin ·
3 3 3 3
 

«1 3 «1 3
= ¬ ’ e12 · 4 e1 ¬ + e12 · = ’2 e1 + 2 3 e 2
¬2 · ¬2 2·
2
  

With the expression of half angle, it is not necessary that the complex number has unity
modulus because:

z ’1 =1’±/2 ¦z¦’1
z =¦z¦1±/2

Then, the rotation through an angle ± can be written as:

v' = z ’1 v z

The composition of two successive rotations implies the product of both complex
numbers, whose argument is the addition of the angles of both rotations.

v'' = 1 ’β / 2 v' 1β / 2 = 1 ’β / 2 1 ’± / 2 v 1± / 2 1β = 1’ ( v 1 (±+β)/2
±+β)/2
/2




Reflections
Figure 4.2
A reflection of a vector in a direction is
the geometric transformation which preserves
the component having this direction and
changes the sign of the perpendicular
component (figure 4.2). The product of
proportional vectors is commutative and that
of orthogonal vectors is anti-commutative.
Because of this, the reflected vector v' may be
obtained as the multiplication of the vector v
by the unitary vector u of the reflection axis at
the left and right hand sides:

with u2 = 1
v' = u v u = u ( v|| + v⊥ ) u = u v|| u + u v⊥ u = v|| ’ v⊥

where v|| and v⊥ are the components of v proportional and perpendicular to u respectively.
Instead of the unitary vector u, any vector d having the axis direction can be
introduced in the expression for reflections whenever we write its inverse at the left side
of the vector:

d vd
= d ’1 v d
v' = 2
d
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 29


Although the reflection does not change the absolute value of the angle between
vectors, it changes its sign. Under reflections, real numbers remain invariant but the
complex numbers become conjugate because a reflection generates a symmetric
parallelogram (figure 4.3) and changes the sign of the imaginary part:

z = a + b e12 a, b real

d ( a + b e12 ) d Figure 4.3
z' = 2
d

d 2 a ’ d 2 b e12
= = a ’ b e12 = z *
2
d

For example, let us calculate the reflection
of the vector 3 e1 + 2 e2 with regard to the
direction e1 ’ e2. The resulting vector will
be:

1 1
( e1 ’ e2 ) ( 3 e1 + 2 e2 ) ( e1 ’ e2 ) = ( e1 ’ e2 ) ( 1 ’ 5 e12 ) = ’ 2 e1 ’ 3 e2
v' =
2 2


Inversions

The inversion of radius r is the geometric transformation which maps every vector
v on r v ’1 , that is, on a vector having the same direction but with a modulus equal to r2 /
2

¦v¦:

v' = r2 v ’1 r real Figure 4.4

This operation is a generalisation of
the inverse of a vector in the geometric
algebra (radius r = 1). It is called inversion
of radius r, because all the vectors with
modulus r, whose heads lie on a
circumference with this radius, remain
invariant (figure 4.4). The vectors whose
heads are placed inside the circle of radius r
transform into vectors having the head
outside and reciprocally.
The inversion transforms complex
numbers into proportional complex numbers with the same argument (figure 4.5):

v' = r2 v ’1 w' = r2 w ’1 z=vw
r4 z
’1 ’1 ’2 ’2
4 4
z' = v' w' = r v w =r vwv w = 2
z
30 RAMON GONZÁLEZ CALVET



Dilatations

The dilatation is that
Figure 4.5
geometric transformation which
enlarges or shortens a vector, that is,
it increases (or decreases) k times the
modulus of any vector preserving its
orientation. The dilatation is simply
the product by a real number k

v' = k v k real

The most transformations of
vectors that will be used in this book
are combinations of these four
elementary transformations. Many
physical laws are invariant under some of these transformations. In geometry, from the
vector transformations we define the transformations of points on the plane, indispensable
for solving geometric problems.


Exercises

4.1 Calculate with geometric algebra what is the composition of a reflection with a
rotation.

4.2 Prove that the composite of two reflections in different directions is a rotation.

4.3 Consider the transformation in which every vector v multiplied by its transformed v' is
equal to a constant complex z2. Resolve it into elementary transformations.

4.4 Apply a rotation of 2π/3 to the vector “3 e1 + 2 e2 and find the resulting vector.

4.5 Find the reflection of the former vector in the direction e1 + e2 .
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 31


SECOND PART: THE GEOMETRY OF THE EUCLIDEAN PLANE


A complete description of the Euclidean plane needs not only directions (given by
vectors) but also positions. Points represent locations on the plane. Then, the plane R2 is
the set of all the points we may draw on a fictitious sheet of infinite extension.
Recall that points are noted with capital Latin letters, vectors and complex
numbers with lowercase Latin letters, and angles with Greek letters. A vector going from
the point P to the point Q will be symbolised by PQ and the line passing through both
points will be distinguished as PQ .


Translations

A translation v is the geometric operation which moves a point P in the direction
and length of the vector v to give the point Q:

+ : R2 — V2 ’ R2

(P, v) ’ Q = P + v

Then the vector v is the oriented segment going from P to Q. The set of points together
with vectors corresponding to translations is called the affine plane1 (R2 , V2, +). From this
equality a vector is defined as a subtraction of two points, usually noted as PQ :

v = Q ’ P = PQ

The sign of addition is suitable for translations, because the composition of translations
results in an addition of their vectors:

Q=P+v R=Q+w R=(P+v)+w=P+(v+w)


5. POINTS AND STRAIGHT LINES

Coordinate systems Figure 5.1

Given an origin of coordinates O,
every point P on the plane is described by the
position vector OP traced from the origin. The
coordinates (c1, c2) of a point are the
components of the postion vector in the chosen
vector base {e1, e2}:

OP = c1 e1 + c2 e2

Then every point is given by means of a pair of

1
The affine plane does not imply a priori an Euclidean or pseudo-Euclidean character.
RAMON GONZALEZ CALVET
32


coordinates:

P = (c1, c2)

For example, the position vector shown in the figure 5.1 is:

OP = 2 e1 + 3 e2

and hence the coordinates of P are (2, 3).
A coordinate system is the set {O; e1, e2}, that is, the origin of coordinates O
together with the base of vectors. The coordinates of a point depend on the coordinate
system to which they belong. If the origin or any base vector is changed, the coordinates
of a point are also modified. For example, let us calculate the coordinates of the points P
and Q in the figure 5.2. Both coordinate systems have the same vector base but the origin
is different:

S = {O; e1, e2} S' = {O'; e1, e2}

Since OP = 2 e1 + 3 e2 and O'P = 2 e2 ,
the coordinates of the point P in the
coordinate systems S and S' are
respectively:

P = (2, 3)S = (0, 2)S'

Analogously:

OQ = ’ 4 e1 + e2 and O'Q = ’ 6 e1
Figure 5.2
Q = (’4, 1)S = (’6, 0)S'

The coordinate axes are the straight lines passing through the origin and having
the direction of the base vectors. All the points lying on a coordinate axis have the other
coordinate equal to zero.
If a point Q is obtained from a point P by the translation v then:


Q=P+v OQ = OP + v

that is, we must add the components of the translation vector v to the coordinates of the
point P in order to obtain the coordinates of the point Q:

(q1, q2) = (p1, p2 ) + (v1, v2)

For example, let us apply the translation v = 3 e1 ’ 5 e2 to the point P = (’6, 7):

Q = P + v = (’6, 7 ) + (3, ’5) = (’3, 2)

On the other hand, given two points, the translation vector having these points as
extremes is found by subtracting their coordinates. For instance, the vector from P = (2, 5)
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 33


to Q = (’3, 4) is:

PQ = Q ’ P = (’3, 4 ) ’ ( 2, 5) = (’5, ’1 ) = ’5 e1 ’ e2

Remember the general rule for operations between points and vectors:

coordinates = coordinates + components

components = coordinates ’ coordinates


Barycentric coordinates

Why is the plane described by complicated concepts such as translations and the
coordinate system instead of using points as fundamental elements? This question was
studied and answered by Möbius in Der Barycentrische Calculus (1827) and Grassmann
in Die Ausdehnungslehre (1844). From the figure 5.3 it follows that:

OR = c1 e1 + c2 e2 = c1 OP + c2 OQ

When writing all vectors as difference of
points and isolating the generic point R,
we obtain:

R = ( 1 ’ c1 ’ c2 ) O + c1 P + c2 Q

That is, a coordinate system is a set of Figure 5.3
three non-aligned points {O, P, Q} so that
every point on the plane is a linear
combination of these three points with
coefficients whose addition is equal to the unity. The coefficients (1 ’ c1 ’ c2 , c1, c2) are
usually called barycentric coordinates, although they only differ from the usual
coordinates in a third dependent coordinate.


Distance between two points and area

The distance between two points is the modulus of the vector going from one
point to the other:

d(P, Q) = ¦PQ¦ = ¦Q ’ P¦

The distance has the following properties:
1) The distance from a point to itself is zero: d(P, P) = ¦PP¦ = 0 and the reverse
assertion: if the distance between two points is null then both points are coincident:

’ ¦PQ¦= 0 ’ Q’P=0 ’
d(P, Q) = 0 P=Q

2) The distance has the symmetrical property:
RAMON GONZALEZ CALVET
34



d(P, Q) = ¦PQ¦ = ¦QP¦ = d(Q, P)

3) The distance fulfils the triangular
inequality: the addition of the lengths of any
two sides of a triangle is always higher than
or equal to the length of the third side (figure
5.4).

d(P, Q) + d(Q, R) ≥ d(P, R)

that is:
Figure 5.4
¦PQ¦ +¦QR¦ ≥ ¦PR¦ = ¦PQ +QR¦

The proof of the triangular inequality is based on the fact that the product of the
modulus of two vectors is always higher than or equal to the inner product:

¦PQ¦ ¦QR¦ ≥ PQ · QR =¦PQ¦¦QR¦ cos ±

Adding the square of both vectors to the left and right hand sides, one has:

PQ2 + QR2 + 2 ¦PQ¦¦QR¦ ≥ PQ2 + QR2 + 2 PQ · QR

(¦PQ¦ + ¦QR¦)2 ≥ (PQ + QR )2

¦PQ¦ + ¦QR¦ ≥ ¦PQ + QR¦

For any kind of coordinate system the distance is calculated through the inner
product:

d2(P, Q) = PQ2 = PQ · PQ

The oriented area of a parallelogram is the outer product of both non parallel sides.
If P, Q and R are three consecutive vertices of the parallelogram, the area is:

A = PQ § QR

Then, the area of the triangle with these vertices is the half of the parallelogram area.

1
PQ § QR
A=
2

If the coordinates are given in the system formed by three non-linear points {O, X,
Y}:

P = ( 1 ’ xP ’ yP ) O + xP X + yP Y

Q = ( 1 ’ xQ ’ yQ ) O + xQ X + yQ Y
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 35



R = ( 1 - xR - yR ) O + xR X + yR Y

it is easy to prove that:

1 ’ xP ’ yP xP yP
PQ § QR = 1 ’ x Q ’ y Q xQ y Q OX § OY
1 ’ xR ’ yR xR yR


Then the absolute value of the area of the triangle PQR is equal to the product of the
absolute value of the determinant of the coordinates and the area of the triangle formed by
the three base points:

1 ’ xP ’ yP xP yP
OX § OY
¦A¦ = ¦ 1 ’ x Q ’ y Q xQ yQ ¦
2
1 ’ xR ’ yR xR yR


Condition of alignment of three points

Three points P, Q, R are said to be aligned, that is, they lie on a line, if the vectors
PQ and PR are proportional, and therefore commute.

” ” PQ § PR = 0 ”
P, Q, R aligned PR = k PQ

PR = PQ ’1 PR PQ
PQ PR ’ PR PQ = 0 ” ”
PQ PR = PR PQ

The first equation yields a proportionality between components of vectors:

yQ ’ y P yR ’ yP
=
xQ ’ x P xR ’ xP

The second equation means that the area of the triangle PQR must be zero:

1 ’ xP ’ yP xP yP
1 ’ xQ ’ yQ xQ yQ = 0
1 ’ xR ’ yR xR yR

that is, the barycentric coordinates of the three points are linearly dependent. According to
the previous chapter, the last equality means that the vector PR remains unchanged after a
reflection in the direction PQ. Obviously, this is only feasible when PR is proportional to
PQ, that is, when the three points are aligned.
RAMON GONZALEZ CALVET
36



Cartesian coordinates

The coordinates (x, y) of any system having an orthonormal base of vectors are
called Cartesian coordinates from Descartes2. The horizontal coordinate x is called
abscissa and the vertical coordinate y ordinate. In this kind of bases, the square of a vector
is equal to the sum of the squares of both components, leading to the following formula
for the distance between two points:

P = (xP, yP) Q = (xQ, yQ)

(x ’ xP ) + ( yQ ’ y P )
d (P, Q ) =
2 2
Q



For example, the triangle with vertices P = (2, 5), Q = (3, 2) and R = (’4, 1) have
the following sides:

¦PQ¦ = 10
PQ = Q ’ P = ( 3, 2 ) ’ ( 2, 5 ) = ( 1, ’3) = e1 ’ 3 e2

50
QR = R ’ Q = (’4, 1 ) ’ ( 3, 2 ) = (’7, ’1 ) = ’7 e1 ’ e2 ¦QR¦ =

40
RP = P ’ R = ( 2, 5 ) ’ (’4, 1 ) = ( 6, 4 ) = 6 e1 + 4 e2 ¦PR¦ =

The area of the triangle is the half of the outer product of any pair of sides:

1 1
PQ § QR = ( e1 ’ 3 e2 ) § (’7 e1 ’ e2 ) = ’11 e12 ¦A¦= 11
A=
2 2


Vectorial and parametric equations of a line

The condition of alignment of three points is the starting point to deduce the
equations of a straight line. A line is determined whether by two points or by a point and a
direction defined by its direction vector, which is not unique because any other
proportional vector will also be a direction vector for this line. Known the direction vector
v and a point P on the line, the vectorial equation gives the generic point R on the line:

r: {v, P} P = (xp, yP) R = (x, y)

PR = k v

Separating coordinates, the parametric equation of the line is obtained:

(x, y) = (xp, yp) + k (v1, v2)

Here the position of a point on the line depends on the real parameter k, usually identified

2
If the x-axis is horizontal and the y-axis vertical, an orthonormal base of vectors is called
canonical.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 37


with the time in physics.
Knowing two points P, Q on the line, a direction vector v can be obtained by
subtraction of both points:

v=Q’P
r: {P, Q}

In this case, the vectorial and parametric equations become:

” R=P+k(Q’P)=P(1’k)+kQ
PR = k PQ

The parameter k indicates in which
proportion the point R is closer to P than to
Q. For example, the midpoint of the
segment PQ can be calculated with k = 1/2.
The figure 5.5 shows where the point R is
located on the line PQ as a function of the
parameter k. In this way, all the points of
the line are mapped to the real numbers
incorporating the order relation and the
topologic properties of this set to the line3.
Figure 5.5
Algebraic equation and distance
from a point to a line

If the point R belongs to the line, the vector PR and the direction vector v are
proportional and commute:

” PR v ’ v PR = 0 ”
PR v = v PR

PR = v ’1 PR v
PR § v = 0 ”

The last equation is the algebraic equation and
shows that the vector PR remains invariant
under a reflection in the direction of the line.
That is, the point R only belongs to the line
when it coincides with the point reflected in
this line. Separating components, the two-point Figure 5.6
equation of the line arises:

x ’ xP y ’ yP x ’ xP y ’ yP
= =

xQ ’ xP yQ ’ yP
v1 v2


3
In the book Geometria Axiomàtica (Institut d'Estudis Catalans, Barcelona, 1993), Agustí
Reventós defines a bijection between the points of a line and the real numbers preserving the order
relation as a simplifying axiom of the foundations of geometry. I use implicitly this axiom in this
book.
RAMON GONZALEZ CALVET
38


where v = ( v1, v2 ).
If the point R does not belong to the line, the reflected point R' differs from R. This
allows to calculate the distance from R to the line as the half of the distance between R
and R' (figure 5.6). The vector going from R to R' is equal to PR' ’ PR .

1 1 1
R' ’ R ¦ = ¦PR' ’ PR¦= ¦ v ’1 PR v ’ PR ¦
¦
d( R, r ) =
2 2 2

We will obtain an easier expression for the distance by extracting the direction vector as
common factor:

PR § v
1 ’1 1 ’1
d( R, r ) = ¦v ( PR v ’ v PR )¦ = ¦v ¦ ¦PR v ’ v PR¦=
2 2 v

In this formula, the distance from R to the
line r is the height of the parallelogram
formed by the direction vector and (figure
5.7).
Similar line equations can also be
deduced using the normal (perpendicular)
vector n. Like for the direction vector, the
normal vector is not unique because any
other proportional vector may also be a Figure 5.7
normal vector. The normal vector
anticommutes with the direction vector of
the line, and therefore with the vector PR,
P being a given point and R a general point on the line:

PR = ’ n ’1 PR n
n PR = ’PR n ” n PR + PR n = 0 ” ”
n · PR = 0

The last equation is also called the algebraic equation and shows that the vector
PR is reversed under a reflection in the perpendicular of the line. Taking components, the
general equation of the line is obtained:

n1 (x ’ xP) + n2 (y ’ yP) = 0 ” n1 x + n2 y + c = 0

where n = (n1, n2) and c is a real constant.
When the point R does not lie on the
line, the point R' reflected of R in the
perpendicular does not belong to the line. Let
PR'' be the reversed of PR', or equivalently
R'' be the symmetric point of R with regard to
P (figure 5.8). Then, the distance from R to
the line r is the half of the distance from R to
R'':

Figure 5.8
1
¦PR'' ’ PR ¦
d( R, r ) =
2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 39


<<

. 2
( 13)



>>