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Appetizers and Lessons for Mathematics
and Reason
New Visitors: Visit site entrance www.whyslopes.com
to see what a 1000+ pages offer.
Yhis webpage offers a short geometric story to introduce the complex
numbers before instead of after trigonometry.
- Step I.
Addition and Multiplication
- Step II. What are Complex
Numbers?
- Step III. Addition and
Multiplication Properties
- Step IV.
Second Way to Compute Products
- Step
V Consequences of Two Ways to Compute Products
The applet shows how to add and multiply
points or vectors in the plane. As the easy consequences, the equality of two ways to
multiply complex numbers simplifies the derivation of trigonometric
identities, gives a simple proof of the cosine, and gives trigonometric formulas
for dot- and cross-products. Details follow below. The circular
trig part of the site area Maps,
Plans, Similarity &Trig, with Complex Numbers
continues the
development of theory of complex numbers with a discussion of roots of unit and
a derivation of of the properties of circular trig functions. (Many
of the references below include similar material)>
- Appendix: Proof of
the Distributive Law - the key element.
In 1976, I saw
Richard Feynman in guest lectures at McGill
University describe the addition and multiplication of vectors in the plane using parallelograms for addition and the rule add
angles, multiply lengths for multiplication. Since then I have wondered
how to transform his presentation. This essay is the result. It
shows how complex numbers may be geometrically introduced with a level of rigour
sufficient for most, before the introduction of periodic trigonometric
functions.
Technical Notes & References
- Teachers: You are welcome to distribute copies or
parts of this document in class, as long as the document source (www.whyslopes.com)
is acknowledged the copies or in each part distributed. Printable
Letter
& Legal
size versions are available in pdf format. Show this
geometric approach to your curriculum or course design and curriculum
committees. The arguments below dispersed over a few high school
courses may accelerate university and university-oriented mathematics
instruction. The presentation below talks about the addition and
multiplication of points in the plane. Instead of points, you could
introduce rectangular polar coordinate description of vectors or arrows in
the plane drawn in standard position, and then talk about coordinate
based addition and multiplication of arrows or vectors drawn in standard
position.
Convention: Square brackets are used to indicate polar coordinates while round brackets
indicate rectangular coordinates.
Addition (Coordinate Method)
The sum of two points with the rectangular coordinates [a,b]
and [c,d] is given by
[a,b] + [c,d] = [a+c,b+d]
Examples:
[2,5]+ [6,2] =
[2+6,5+2] = [8,7].
[-1,14]+ [2,8] = [-1+2,14+8] = [1,22].
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Optional Readings: For
points [a,b], [c,d] and
[e,f] in the plane, addition is both commutative and
associative:
The commutative property
[a,b]+[c,d] = [c,d]+[a,b]
holds as a+ c = c + a and b + d = d + b due to commutative
property of the
addition of real numbers.
The associative property
( [a,b]+[c,d] ) +[e,f]
= [a,b]+( [c,d]) +[e,f] )
holds as
(a+ c) + e = a + ( c + f) and (b +d) + f = b + (d +f)
due to the associative law for real numbers |
If point A = [a,b] in rectangular coordinates,
then A = (r, q
) in polar coordinates where r is the distance of A to the origin,
q is angle. When A = [a,b] and A = (r, q
), we write [a,b] = (r, q
). Both rectangular and polar coordinates may be use to locate a
point in the plane.
The product of two points [x1, y1] = (r1,q1)
and [x2,y2] = (r2,q2)
in the plane is calculated using the polar coordinates with the multiply the lengths,
add
the angles rule:
[x1, y1]·[x2,y2]
= (r1,q1)·(r2,q2)
= (r1r2,q1+q2)
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Example. Two arrows are to be multiplied. One has length 1.3 and angle
22.62°; the other factor has length 1.026
and angle 46.97°; and so their product
has length 1.3338 = 1.3·1.026 and angle 69.59°
= 22.62°+46.97°;
and that is it. See the following diagram.

correction: 22.62 + 46.97 = 69.59 not 69.69s
Another Example. The product of the two points (3,80°)
and (4, 60°) is
(3 . 4, 80°+ 60°)
= (12,140°)
The product is a third point [x3,y3] = (r3,q3).
Its rectangular coordinate [x3,y3] are determined by
the values of with r3 = r1r2
and q3 = q1+q2
. Below, we will see how to compute the product using the rectangular
coordinates of the factors [x1, y1] and [x2,y2]
directly, without use of polar coordinates. |
Optional Readings:
Commutative Law for Products of points in the plane:
(r1,q1)·(r2,q2)
= (r2,q2)·
(r1,q1)
whenever (r1,q1)
and (r2,q2)
are polar coordinates for a pair of points in the
plane. This property follows as the commutative law
of addition for real numbers (or angles >
0) implies q1+q2
= q2+q1
and the commutative property for products of real numbers (>
0) implies r1r2 = r2r1,.Commutative
Property for Products of points in the plane:
{(r1,q1)·(r2,q2)}·
(r3,q3)
= (r1,q1)·{(r2,q2)·
(r3,q3)}
whenever (r1,q1),
(r2,q2)}and
(r2,q2)
are polar coordinates for a pair of points in the
plane. This property follows as the associative property of addition for real numbers (or angles >
0) implies {q1+q2}+
q3= q1+
{q2+ q3}
and the commutative property for products of real numbers (>
0) implies {r1r2}r3
= r1 {r2r3 }
Multiplicative Inverse for non-zero points in
the plane.
(r1,q1)·(
[1/r1] , -q1)
= (1,0)
if (r1,q1)
is the polar coordinates of a nonzero point in the plane. |
Recapitulation - Summary
The addition of points in the plane is given by means of
their rectangular coordinates while multiplication is given
in terms of polar coordinates. Below you will see how to
multiply points together using rectangular coordinates as
well. The equality of different ways to multiply points
together leads to many properties of vectors and
trigonometry.
The following applet shows how to
multiply and add points or vectors in the plane.
Help function button will take you back to this
page. This page is the help file as it introduces complex numbers.
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Help function button will take you back to this
page. This page is the help file in that it explains complex numbers.
For multiplication, add the angles and multiply the lengths to get
the product. Hit the button C=A*B twice to turn on and off the arc
illustration of angle addition. The illustration works best when both
factors are in the first quadrants. Hit the button C=A+B to again and
again to alternate between illustrations of the head-to-tail method and
the component method for vector addition. Now experiment or play.
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Points in the plane with the operations of addition and multiplication just
given are called the complex numbers. The plane with these two operations on its
points is called the complex numbers plane, or more briefly the complex numbers.
We will now change to a more standard notation
for them. We may and often will write the rectangular coordinates z = [a,b]
as z
= a+ib, We will further call the abscissa a, the real part
of the complex number z = a+ib. We will also call the
ordinate b, the imaginary part of the complex number z = a+ib.
Note the previous notations [a, b] and (r,q) will
be used for points in the plane in the further discussion of the
properties of real and complex numbers. Eventually, the previous notation [a, b] and
(r,q) will
be phased out.
Purely Imaginary: We will say that the complex number z = a+ib
is purely imaginary when and only when its real part a = 0. The angle
of a purely imaginary complex number z = a+ib = 0+ib
= (0,b) is 90 degrees or 270 degrees (modulo 360 degrees). When b
> 0, the angle is 90 degrees (modulo 360 degrees). When b < 0, the
angle is 270 degrees (modulo 360 degrees).
Note: Two quantities x and y are equal modulo a third
quantity c, if and only if their difference x-y
=
kc for some whole number or integer k.
Real: We will also say that z = a+ib is (purely)
real when and only when its imaginary part b is zero. The angle of a
(purely) real complex number z = a+ib = a+i0
= (a,0) is 0 degrees or 180 degrees (modulo 360 degrees), depending on
the sign of the real part a. If a > 0, this angle is 0 degrees
(modulo 360 degrees) while if a > 0, this angle is 180 degrees (modulo
360 degrees).
Each complex number z = a+i0 with imaginary part zero
gives and is given by a real number a. We will write z = a
in this situation, and say that the complex number z is also a real number.
With this practice, the real numbers can be regarded as a subset of the
complex numbers; and the real number line can be identified with the
horizontal axis of the plane.
Real and Imaginary Parts
For each point or complex number z = a+b i = (a,b)
= [r,q] in this plane, we say that a is
the real part of z; that b is the imaginary part of z;
that r = |z| is
the magnitude, modulus or absolute value of z (different texts
prefer different terms); and that q is the angle or argument
of z.
Exercise: Use b = sign(b)|b| to show that bi = b.
i where i = [0,1]
We identify the real number line with the horizontal axis of the plane. With
this identification, observe that positive numbers have angular displacement
zero, modulo 360 degrees. Also observe that negative numbers have angular
displacement 180 degrees, modulo 360 degrees. The magnitude of a real number is
its distance to the origin.
Suppose z = a+i0 and w = c+i0. We
want to compute the product zw with the multiply the lengths, add the
angles rule. Each factor has length |a|
or |c|. Each factor
has angle 0 or 180 degrees (modulo 360 degrees). The relationships
- 0° = 0°+0°
- 180° = 0°+180°
= 180°+0°
- 360° = 180°+180°
= 0° (modulo 360°)
imply the add the angles, multiply the lengths rule for the
multiplication of complex numbers agrees with the ordinary method for
multiplying real numbers and the law of signs. The relationship in particular
imply
- (+1) = (+1)(+1) as 0° = 0°+0°
- (-1) = (+1)(-1) = (-1)(+1)
as 180° = 0°+180°
= 180°+0°
- (-1)(-1) = (+1) as
360° = 180°+180°
Examples and then some further comments may reinforce these ideas. For the first
example, the number 4 is now identified with the point (4,0) = [4,0°]
= [4,360°]. This number or point has
distance 4 to the origin and angle of 0°,
modulo 360 degrees, with the horizontal axis:
For the second example, the number -2
is identified with the point [-2,0] = (2,180°).
See the figure below.

Now multiplying the point (2,180°) by
itself leads to the product (2,180°)2
= (22,180°+180°)
= (4,360°) = (4,0°).
Thus the point on the horizontal axis identified with -2
when squared gives the point identified with +4 indicated above. The 360 degrees
in the diagram for the number or point 4 = [4,0] represents the doubling of the
angle 180 degrees.
For an example or exercise, compute the pair-wise products of 3=3+0i, 4=4+0i,
-3=-3+i0 and -4=-4+0i using the add the angles, multiply the lengths rule.
More Exercises. Compute the following using the multiply the lengths,
add the angles rule:
- A = (1.5)·(2).
- B = (1.5)·(-2).
- C = (-1.5)·(-2).
- D = (1.5)·(-2).
- E = (10,45°) ·(1/20,15°).
Note each factor gives a point or arrow in the coordinate plane.
Stop For A Summary. The polar coordinate definition
(r1,q1)·(r2,q2)
= (r1r2,q1+q2)
of the product of two point in the plane, involves the multiplication of lengths
(= distances to the origin) and the addition of angles. For points on the
horizontal axis, the angles of the factors are zero or 180°
(modulo 360°). Computing the angle of the
product will involve one of the following expressions:
0°+0°
= 0°
0°+180°
= 180°
180°+0°
= 180°
180°+180°
= 360°
since the angle 180 degrees is associated with -1, and the
angles 0 and 360 degrees are both associated with the number +1, the polar
coordinate definition of multiplication of points in the plane agrees with (or
yields) the law of signs for the multiplication of positive and negative
numbers.
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The real number -1 = -1+0i
= (1,180°) has angle 180 degrees (mod 360
degrees) and length 1. The purely imaginary number [0,1] = 0+i1 = (1,90°)
has angle 90 degrees and length 1. Multiplying this point or number by itself,
that is, squaring it, gives the point with length 1 ×1 = 1 and angle 90°+90°
= 180°. So the product equals -1+0i
= -1. We call i, the principal square
root of -1.
A second square root of -1 is obtained as follows. The imaginary number (0,-1)
= 0+i(-1) = [1,-90°]
has angle -90 degrees and length 1. Multiplying this
point or number by itself, that is squaring it, gives the point with length 1
times 1 =1 and angle (-90°)+(-90°)
= -180° = 180°
(mod 360°). So this product equals -1+0i
= -1 as well.
This provides two square roots of -1 as both (1,+90°)2
= (1,+180°) = -1
and (1,-90°)2
= (1,-180°)
= -1.
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The square root of a positive number or zero are real nonnegative numbers. I
assume in the following that you know how to compute these square roots. The
square roots of negative numbers and of other arrows or points in the coordinate
plane depend on this ability.
Observe that squaring points in the plane doubles their angular displacements
and squares their magnitudes (distance to the origin). That is, the add the
angles, multiple the lengths rule gives
[r½, ½ q]·[r½,½
q]
= [r ,q]
Therefore the arrow (r½, ½q) when
squared (meaning multiplied by itself) yields (r,q)
. So it is called a square root of the arrow (r,q).
Another square root is located by the polar coordinates (r½, ½q+180°)
since (r,q) = (r,q+360°)
both locate the same point in the plane. You should consider the special case of
positive numbers z = a+i0 = (a,0°)
where the angle q = 0 degrees.
Exercises.
- Find all the square roots of 4 and -4 and plot them.
- Find the cube roots of 27 and -27 and plot them in the plane.
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The complex conjugate of a complex number z = a+b i
with polar coordinates (r, q) is the complex
number `z = a-b
i with polar coordinates (r, -q). Multiplying
a complex number a+b i by its conjugate a-bi
gives the nonnegative number r2 > 0
Observe that p = [(a)/(r2)]-i[(b)/(r2)]
= [1/(r2)][`(z)] has angle -q
and length [1/(r)]. Here p = [1/(r2)][r,-q]
= (1/r,-q.) Multiplying number p =
[[1/(r)],-q] by z = [r,q]
gives the complex number [1,0] with length 1 and angle 0, that is, the real
number 1. And multiplication of any point (c,d) by 1 = [1,0°]
yields back the point (c,d)
The reciprocal (or multiplicative inverse) of the complex number z = a+b
i with length r > 0 and angle q is
the complex number p with length 1/r and angle -q.
Observe that if r > 1 then the length of the
reciprocal [1/(r)] < 1 < r, that is, the length of the
reciprocal is less than 1 and the length of the original number. In contrast, if
0 < r < 1 then [1/(r)] > 1 > r. Question: Which
of these two cases is represented in the above diagram? What happens in the case
r = 1?
Some Vocabulary.
For each point or complex number z = a+b i = (a,b)
= [r,q] in this plane, we say that a is
the real part of z; that b is the imaginary part of z;
that r = |z|
= Ö[(a2+b2)] is
the magnitude, modulus or absolute value of z (different
texts prefer different terms); and that q is the
angle or argument of z.
Below z, w and v stand denote complex
numbers. The following properties are consequences of corresponding
properties of real numbers and the rectangular and polar coordinate methods for
calculating sums and products of points in the plane.
- Commutative Property for Addition:
z + w = w+z
- Commutative Property for multiplication:
zw = wz
- Additive Identity Exists: The zero vector
0 = 0 + i 0 has the property 0 + z = z
- Multiplicative Identity Exist: The real
number 1 = 1 + i 0 has length 1 and angle 0. So it has the property
that 1 z = z.
- Reciprocals (Multiplicative Inverses) Exist for
nonzero complex numbers: If z = (r, -q) has
length r> 0 and angle q then
wz = 1 if w = (1/r, -q)
with length (1/r) and angle -q.
- Negatives (Additive Inverses) Exist for all complex numbers:
If z = a + ib = [a, b] then w = (-a) + i(-b) = [-a, -b] has the
property that w+z = 0
- Non Zero Product Property: If z and w have nonz-zero
magnitudes (lengths) r and s then their product has magnitude or length rs > 0.
So the product is nonzero
From logic,
the equivalent, contrapositive form of the nonzero product law is as follows:
Zero Product Law: If the product wz = 0 for a pair of complex
numbers factors z and w then at least one of the factors must be
zero.
The Distributive property says
Z ( W + V ) = Z W + ZV
(left distributive law)
( W + V ) Z = WZ + V Z (right distributive law)
The appendix below
(optional reading) provides a proof. The two laws are equivalent due to
the commutative property of multiplication.
The distributive law applied twice implies
(a+ib)(c+id) = (ac-bd)+i(bc+ad)
for the product of two points in the plane in terms of their rectangular
coordinates, alias real and imaginary parts.
Proof:
(a + bi) (c+ di) = a(c+di) + bi (c+ di)
(by first use of distributive law)
= ac+ a(di) + (bi)c+ (bi)(di)
(by second use of
distrributive law)
= ac + i ad + i bc + (-1) bd
( by associative and
commutative law for products)
= ac + (-1) bd + i ad
+ i bc
(by associative and
commutative laws for sums)
= 1 (ac + (-1) bd) + i (ad
+ bc)
(by the distributive law
in reverse)
= [ac + (-1) bd , ad + bc]
The foregoing gives a second way to multiply complex numbers together
using their real and imaginary parts
(a + bi) (c+ di) = (ac - bd) + i (ad + bc)
or equivalently, with or rectangular coordinates notation,
[a,b] [c,d] = [ac -bd, ad+ bc]
The latter formulas often the starting point for the definition of
products of complex numbers before the introduction of complex number
notation in the plane.
Exercise: Use b = sign(b)|b| to show that bi = b.
i where i = [0,1]
Mathematicians, engineers and physicist know well how the properties of
complex numbers and the function cis(q) = cos(q)
+ i sin(q) = eiq
simplifies the development of trigonometry identities. The following easy
consequences are likely to be less well-known.
From trigonometry, recall the unit circle definitions of the sine and cosine
functions. Let
e iq =
cis(q) = cos(q)+isin(q)
= exp(iq) = [cos(q),
sin(q)] = (1, q)
Now (1,A)·(1B) = (1, A+B). Therefore property
cis(A)·cis(B) = cis(A+B)
follows from the above add the angles, multiply the lengths definition
of complex multiplication.
From cis(A) = cos(A)+isin(A)
= a + bi, and cis(A) = cos(B)+isin(B) =
c + id
cis(A+B) = cis(A)·cis(B)
= (a+ib)(c+id) = (ac-bd)+i(bc+ad)
= { cos(A)cos(B) -sin(A)sin(B) } + i{sin(A)cos(B)+ cos(A)sin(B)}
But cos(A+) = cos(A+B) + i sin(A+B) as well. Therefore in
rectangular coordinates
[cos(A+B), sin(A+B)] = [ cos(A)cos(B) -sin(A)sin(B),
sin(A)cos(B)+ cos(A)sin(B)]
Equality requires the angle sum formulas to hold.
cos(A+B) = cos(A)cos(B) -sin(A)sin(B)
sin(A+B) = sin(A)cos(B)+ cos(A)sin(B)
The verification or derivation of trig identities can be reduced to algebraic
manipulations involving
cis(q) = cos(q)
+ i sin(q) = exp(i q)
Suppose [x1,y1] = [ r1 cos(q1),
r1 sin(q1)] and [x2,y2]
= [ r2 cos(q2), r2
sin(q2)] are points in the plane. Then
their dot product
[x1,y1].[x2,y2]
= x1x2+y1y2 (dot product
definition)
and their cross product
[x1,y1].[x2,y2]
= x1y2 - y1x2
(cross product definition)
may be expressed in terms of trigonometric functions and the angles
between the two points, or more precisely their position vectors. See below.
Details
To obtain the geometric interpretation, observe the polar and
rectangular ways to multiply the first point by the complex complex conjugate of
the second, when both are viewed as complex numbers: That is,
[x1,y1][x2, -y2]
= (r1 , q1)(r2, -q2)
From the equality of two different ways to multiply points in the plane,
observe
[x1x2+y1y2 , x1y2
- y1x2] = (r1r2,q1-
q2)
but
(r1r2,q1-
q2) = [ r1r2
cos( q1- q2),
r1r2 sin( q1-
q2)]
Therefore comparison (equality) of real and imaginary parts
yields:
x1x2+y1y2 = r1r2
cos(q)
and
x1y2 - y1x2
= r1r2 sin( q)
where
q = q1-
q2
is the angle between the two points [x1,y1]
= (r1 , q1) and [x2,y2]
= ( r2 , q2)

In the development of complex numbers, we may geometrically imply that the
product (r,0)·[a,b] = [ra, rb] = r·[a,b] is given by a scalar multiplication,
and that the midpoint of the line segment from [a,b] to [c, d] is ½·[a+c,b+d]
via similarity and triangle isometry (congruency) arguments. With that
the properties of complex numbers employed are not derived from the Pythagorean
theorem.
Theorem: If a right triangle has a hypotenuse of length r and other two
sides of lengths a and b,
then
a2+b2 = r2
Complex Number Proof
Let z = a + i b = (r, A) be a point in the first quadrant. The
triangle with vertices 0, a, a+ ib is congruent or isometric to the given right
triangle

Multiplying a vector a + i b with angle A and length r by its complex
conjugate a - ib gives a complex number with angle A + (-A) = 0 and
length r2 units according to the add the angles, multiply the
lengths polar coordinate, multiplication rule. The product has value r2
> 0 as shown below.

This gives
r2 = (a+ib) (a-ib)
But the previous formulas for expressing products in terms of their
real and imaginary parts,
r2 = a2 + b2 + i(-ba+ab)
as the polar and rectangular methods for computing a product give the same
result.
This implies
r2 = a2 + b2
Small Print: There are over 100 different proofs of the Pythagorean
theorem. Using the Chinese square dissection method to imply the
Pythagorean theorem, or another method to imply the Pythagorean theorem and then
use the latter instead of triangle similarity and isometric arguments to imply
that the product (r,0)·[a,b] = [ra, rb] = r·[a,b] is given by a scalar
multiplication, and that the midpoint of the line segment from [a,b] to [c, d]
is ½·[a+c,b+d] gives an easier route for the exposition of complex
numbers. Whence the above arguments simplify shows the Pythagorean theorem
is consistent with the geometric properties of complex numbers.
For a triangle, with sides of length a, b and c, and angle q
opposite the side of length c,

the cosine law say c2 = a2 + b2
- 2ab cos(q).
Proof of the Cosine Law
Introduce a coordinate system so that the origin z = 0 is at the
angle and the ends of the adjacent sides have coordinates [x1,y1]
and [x2,y2]. Then the following diagram and
computation of the dot product implies

and computation of the dot product implies
x1x2+y1y2 = r1r2
cos( q) = ab cos( q)
But
c2 = (x1 - x2)2
+ (y1 - y2)2 = (x12
- 2x1x2 + x22 ) +
(y12 - 2y1y2 +
y22 )
= (x12 + y12 )
+ (x22 + y22
) - 2(x1x2 + y1y2
)
= a2 + b2 - 2ab cos( q)
If the sides of a triangle have length a, b and c with c2
= a2 + b2 then the triangle has a right angle opposite the
side of length c.

Proof of Converse:
When 0 < q < 180,
cos(q) = 0 when and only when q
= 90 degrees
Therefore c2 = a2 + b2 - 2ab cos(q)
= a2 + b2 with a > 0, b >0 and 0 < q
< 180 implies cos(q) = 0 and hence q
= 90 degrees. So the triangle is a right triangle.
Formulas for cos(2A) and sin (2A)
From cis(2A) = cis(A)*cis(A) we found
cos(2A)+ i sin(2A) = (cos (A) + i sin(A)) (cos (A) + i sin(A))
= cos2(A)-sin2(A) + i 2cos(A)sin(A)
Therefore comparison of rectangular coordinates/components (or real and
imaginary parts) yields the double angle formulas
cos(2A) = cos2(A)-sin2(A)
sin(2A) = 2 i cos(A)sin(A)
Formulas for cos(nA) and sin (nA), the case n =3.
Observe
exp(i3A) = cos(3A) + i sin (3A) = exp(iA) exp(iA)
exp(iA)
Therefore
exp(i3A) = {cos(A) + i sin (A)}3
= .cos3(A) + 3cos2(A) i
sin(A) + 3 cos(A) {i sin(A)}2 + (i sin(A))3
= .cos3(A) - 3 cos(A) sin2(A)
+ i{3cos2(A) sin(A) - sin3(A))
Equality of corresponding real and imaginary parts gives
sin (3A) = 3cos2(A) sin(A) - sin3(A) = 3( 1- sin2(A))sin(A)
- sin3(A)
= 3 sin(A) - 4 sin3(A)
cos(3A) = .cos3(A) - 3 cos(A) sin2(A) =
cos3(A) - 3 cos(A) (1-cos2(A))
= 4 cos3(A) - 3 cos(A)
Exponential or cis Expressions for Trig Functions
From the two equations
exp(iA) = [cos(A),sin(A)] = cos(A) + i sin(A)
exp(-iA) = [cos(-A),sin(-A)] = cos(A) - i sin(A)
we see
exp(iA) + exp(-iA) = 2 cos(A)
and
exp(iA) - exp(-iA) = 2i sin(A)
Therefore
(1/2)[exp(iA) + exp(-iA)] = cos(A)
and
(1/2i) [exp(iA) - exp(-iA)] = sin(A)
Expressions involving exp(iA) and exp(-iA) now follow from:
tan(A) = sin(A)/cos(A)
So all trig functions may be expressed in terms of exp(iA) and exp(-iA).
The substitution of exp(iA) and exp(-iA) expressions for trig functions turns
the proof or derivation of trig identities into simple algebraic exercises
involving complex numbers and the add the angles rule for multiplication of
exp(iA) with exp(iB). While highschool students may be
taken through the exercises of proving trig identities before meeting complex
numbers, the above quick explanation of complex numbers and its links to
trig imply a quick route through high school mathematics courses on algebra and
trig. In retrospect, the presentation of trig in highschool has been
harder than need-be.
End Notes
Trig course today could cover the above material, show how most trig
identities follow from calculations with complex numbers, and give applications
of trigonometry to distance calculations based on the similarity of right
triangles and the values of trigonometric functions. A course on trig and
complex numbers could explore more analytic geometry, show how to compute
powers and roots for positive real numbers using the natural logarithm (defined
for positive numbers) and exponential functions (defined for real numbers), and
then extend these definitions to give definitions of powers and roots for
complex numbers, including negative real numbers. Calculations of roots of
unity would further tie trigonometry and complex numbers together. More
motivation and more applications of trig, tied to right triangles and complex
numbers, come from engineering and physics. The description of
electric currents and devices depends on sine and cosines, or more simply, if
you know complex numbers and exp(iA) = exp(iA). The latter functions also appear
in the theoretical treatment of statistics and of higher mathematics (analysis).
-
Mathematical Thought from Ancient to Modern Times, by Morris
Kline, as three volumes (1990, published by Oxford University
Press).
In volume 2, Chapter 27, the
third section called The Geometrical Representation of Complex Numbers briefly describes the approach of Caspar Wessel (1745-1818). Part of
Wessel's work (translated into English) is reproduced in David Eugene
Smith's 1929 work A Source Book in Mathematics, Dover 1959 Reprint.
- What is Mathematics, R. Courant & H. Robbins, Oxford University
Press, Fourth Edition.
Classic Work. This may be taken a prequel to the discussion in the
1950s of what should be taught in pre-university mathematics. Very
readable for undergraduate students in mathematics. The geometric interpretation (or representation) of complex numbers assumes
the addition theorems (angle sum formulas) for sine and cosines in order to
show how to multiply complex numbers using moduli and angles.
- Secondary Mathematics, A Functional Approach for Teachers, H. F.
Fehr, D. C Heath and Company Boston 1951.
The chapter pp254-296 on complex number systems and trigonometry gives as a
exercise for students (!) the task of giving a geometric proof of the
distributive law for complex numbers when multiplication is defined by
multiplying moduli and adding angles. This site December 2009
proof below give the most recent and simplest site proof of the distributive
law, the simple proof I have looking for since seeing Feynman in 1976
describe physics in terms of adding and multiplying vectors in the
plane. I should have known about Fehr work earlier. My copy was a
fall 2009 gift from a McGill University colleague.
-
A History of Algebra from al-Khwarizmi to Emmy Noether, B. L.
van der Waerden. Springer Verlag, ISBN 3-540-13610-X, 260+
pages.
Page 178 says the following regarding complex numbers: Euler
... did not give a satisfactory definition. Clear, geometrical
definitions ... were given by Caspar Wessel in 1997, by Jean Robert Argand
in 1806, by John Warren in 1828, and by Carl Fredrick Gauss in
1831. ...William Rowen Hamilton defined (1843) the complex
numbers as pairs of real numbers subject to ... rules of addition and
multiplication. Augustin Cauchy interpreted (1847) the complex numbers as
residue classes of polynomials,..., modolo x2 +1
Let P and Q be points in the plane. The midpoint of P and Q is M = ½
·(P+Q) = ½· S where S is the sum of the two points

Figure 1: Two Points P and Q in the Plane and
the origin determine a triangle POQ
Multiplication by (r,q) =(1 ,q
)·(r ,0) is the same as multiplication by a unit
length rotation (1 ,q ) and a zero
angle (r,0) point. But multiplying by a zero angle point
is equivalent to the scalar multiplication r [a,b] = [ra,rb], and that
is distributive.
In general, save for a collinear case to be treated separately, the
two points P = [a,b]=(r1,q1)
and Q= [c,d] = (r2,q2)
along with the origin [0,0] are the vertices of a triangle The midpoint
M of the side opposite the origin has rectangular coordinates ½ { [a,b]
+ [c,d]}= ½ { P + Q }= ½ S. But multiplication by
(1 ,q ) rotations the vertices into P' =
(1 ,q )·[a,b] = (r1,q
+ q1),
(1 ,q )·[c,d] = (r2,q+q2),
and Q' = (1 ,q )·[0,0] = [0,0].
The latter provide vertices of a second triangle.
Next we show that rotation by angle q
Distributes over Midpoint Calculation. Details follow.

Scalar multiplication by 2 is equivalent to multiplication by the zero
angle point (2,0) and multiplication commutes. A scalar multiplication by 2
now implies the distributive law for multiplication by (1 ,q
).
P' + Q' = 2 M' = 2 (1 ,q )·M
= (1 ,q )·2M = (1 ,q
) (P + Q)
Therefore
(1 ,q )·P +
(1 ,q )·Q = (1 ,q
)·(P + Q)
Multiplication by (r,q) =(1 ,q
)·(r ,0) being equivalent to two successive
distributive operations is also distributive.
The case where the origin O =[0,0], and points P and Q are collinear follow
from similar or easier arguments in the two subcases where P and Q are on (i)
the same side and (ii) opposite sides of the origin.
Variation of Proof: If we were to take P = [2a,2b] =(2r1,q1)
and Q = [2c, 2d] =(2r2,q2)
above then the midpoint M = ½ ·{P + Q} = [a,b] + [c,d]. Further
the conclusion (1 ,q )·M = M'
would follow as above. That implies
(1 ,q )·{[a,b] + [c,d]} = (1 ,q )·M
= M'
= ½ ·{P' + Q'}
= ½ ·{ (1 ,q )· 2 [a,b] + (1 ,q )·
2[c,d]}
= (1 ,q )·[a,b] + (1 ,q )·
[c,d]}
Remark for Students of Differential Geometry: The observation
that the sum of two points [a,b] and [c,d] in the plane is the midpoint of the
line segment joining [2a,2b]=(2r1,q1)
and [2c, 2d] =(2r2,q2)
provides a geometric alternative to the parallelogram method for obtaining the
sum. On Riemann surface, choose a point O to serve as the origin of a
geodesic based polar coordinate system (r ,q ).
Locally, let the (symmetric) sum S of two points (r1,q1)
and (r2,q2) be given
by the midpoint of the geodesic joining (2r1,q1)
and (2r2,q2). And in that polar coordinate system, for each positive number K, the
K-product is given by (r1,q1)·(r2,q2)
= (Kr1r2,q1+q2)
for some Thus two operations may be locally defined in a neighborhood of
point on a Riemann surface. If the polar disk r1 <
1/K lies in that neighborhood, it is closed under the K-product
operation. The circle of radius 1/K is invariant When the Riemann surface is rotational symmetric about the
origin of the polar system, as in the case of the sphere or plane, rotations
distribute over addition but dilations c (r ,q )
= (cr ,q ) on the sphere do not.
Question: For the addition of points a and c on the real number
line, let a + c be the midpoint for the line segment with vertices 2a
and 2c. Clearly a + c = c + a. Is it is possible for that
introduction of addition of points on the real number line to aid the
development of arithmetic with whole numbers and fractions or integers
and rational numbers?
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