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Location: Site Entrance < Volume 3 Why Slopes - A Calculus Intro Etc << Chapter 23 Links To Trigonometry

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Chapter 23. Complex Numbers - Links to Trig

Volume 3, Why Slopes and More Math.

Note- December 2010: The year 2010 Postscripts 1 to 4 make some chapter content (written 194-96) partially redundant.

The addition of complex numbers or points in the plane was given by means of their rectangular coordinates while multiplication was given in terms of polar coordinates. It is still an exercise perhaps in trig, an application of the angle sum formulas, to obtain expressions for the rectangular coordinates of a product in terms of the rectangular coordinates of the factors. Another exercise or alternative, is to justify and then apply the distributive law for complex multiplication over addition. The application bypasses the exercise or link with trig just indicated in favor some algebraic manipulation. The alternatives are given. Which approach to favor may be a matter of taste. This chapter assumes you are familiar with the unit circle definition of sines and cosines and with the addition of vectors.

The cis or exponential functions

From trigonometry, recall the unit circle definitions of the sine and cosine functions. Let
cis(q) = cos(q)+isin(q) = exp(iq)
of a purely imaginary argument. It is now easy to say how and why the property
cis(A)路cis(B) = cis(A+B)
follows immediately from the above add the angles, multiply the lengths definition of complex multiplication. Hint: both factors have unit lengths.

Applying the Angle Sum Formula

Note the angle sum formulas for cosine and sine met in this section will be proven later. When a+bi = (a,b) = [R,q] and c+id = (c,d) = [r,b] basic trigonometry gives
a
=
R cos(q)
b
=
R sin(q)
c
=
r cos(b)
d
=
r sin(b)

Now the add the angles, multiples the length rule implies the product (a+bi)(c+id) has polar coordinates [Rr,q+b] = (x,y). The rectangular coordinates (x,y) of the product are therefore given by

x
=
Rr cos(q+b) and
y
=
Rr sin(q+b)
But the angle sum formula for cosine say
cos(q+b) = cos(q)cos(b)-sin(q)sin(b)
This implies the real part
x
=
Rr cos(q+b)
=
Rr (cos(q)cos(b)-sin(q)sin(b))
=
(R cos(q))(r cos(b))-(R sin(q))(r sin(b))
=
ad-bc
This expresses the real part x of the product x+iy = (x,y) = (a+bi)(c+id) in terms of the real and imaginary parts of the factors, that is their rectangular coordinates. The argument for the imaginary part is similar. It is given next.

The angular sum formula for sine says

sin(q+b) = sin(q)cos(b)+cos(q)sin(b)
This implies the imaginary part
y
=
Rr sin(q+b)
=
Rr (sin(q)cos(b)+cos(q)sin(b))
=
(R sin(q))(r cos(b))+(R cos(q))(r sin(b))
=
bc+ad
This expresses the imaginary part y of the product x+iy = (x,y) = (a+bi)(c+id) in terms of the real and imaginary parts of the factors, that is their rectangular coordinates.

Product Rule in terms of Rectangular Coordinates

The foregoing yields the expression

(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.

Properties of Complex Numbers

The assumption that the addition and multiplication of positive real numbers a, b, and c are commutative and associative operations implies the following. The length and angle defined multiplication of complex numbers is a commutative and associative operation as well. Details are omitted. They are left as an exercise.

Another Exercise. Employ the expresson (a+ib)(c+id) = (ac-bd)+i(bc+ad) to show that the distributive law for real numbers and the above expressions for the real and imaginary part of the product of two complex numbers implies the distributive property a(b+c) = ab+ac holds for any triplet a = a1+ia2, b = b1+ib2 and c = c1+ic2 of complex numbers.

Trig Identities Simplified

The verification or derivation of trig identities can be reduced to algebraic manipulations involving the cis(q) function

cis(q) = cos(q) + i sin(q) = exp(i q)
and the property
cis(a) cis(b) = cis(a+b)
exp(ia) exp(ib) = exp(i(a+b))
The latter property, as said, follows from the polar coordinate, add the angle, multiply the lengths rule for multiplication of complex numbers.

The demonstration of many trig identities without the use of the properties of the function cis(iq) is an enormous task, or a delicate procedure, a needless but traditional exercise in some trigonometric courses.

BabyTalk: Trig students meet the cis function (cosine i sine function) when teachers or course design prefer not to speak about exponentials exp(i q) of imaginary or complex numbers.

Angle Sum Formulas, Two Proofs

Two proofs of the angle sum formulas are given next. The first recalls the rotate-a-triangle proof met in trigonometry. The second derives the formulas in three steps from the add the angles, multiple the lengths definition of complex multiplication. Both proofs rely on the addition of angles and rotation of points on the unit circle. The rest of this section is optional reading especially if you have seen proofs of these angle sum formulas.

First Proof

This proof of the angle sum formula for sines and cosines involves the rotation about the origin by the angle b of a triangle formed by the vertices (0,0), (1,0) and (cos(a-b),sin(a-b)) into the triangle with vertices (0,0), (cos(b),sin(b)) and (cos(a),sin(a)) respectively. We assume this rotation does not change the lengths of the sides of the triangle - that it is, a rigid body motion.


The original triangle and the rotated triangle both have a side that is a chord on the unit circle. The length of this side is not changed by the rotation, at least that is our geometric assumption. Therefore computation of chords length squared can be done with the help of its end-point coordinates before and after rotation. And the two computations should give the same result. This implies

cos(a-b)-1)2+(sin(a-b)-0)2 = (cos(a)-cos(b))2+(sin(a)-sin(b))2

The latter in turn implies
1-2cos(a-b)+(cos(a-b))2+(sin(a-b))2
=
(cos(a))2-2cos(a)cos(b)+(cos(b))2
+(sin(a))2-2sin(a)sin(b)+(sin(b))2

Therefore

2-2cos(a-b)
=
2-2cos(a)cos(b) -2sin(a)sin(b)
From this,
cos(a-b) = cos(a)cos(b) +sin(a)sin(b)
The replacement of b by -g yields
cos(a+g) = cos(a)cos(g)-sin(a)sin(g)
as sin(-g) = -sin(g). The identity cos(q) = sin(90掳-q) implies the identity cos(90掳-q) = sin(q). Together, the identities imply
sin(a+g)
=
cos(90掳-a-g)
=
cos(90掳-a)cos(-g)- sin(90掳-a)sin(-g)
=
sin(a)cos(-g)- cosa)(-sin(g))
=
sin(a)cos(g)+ cosa)(sin(g))
as cos(-q) = cos(q). This completes the rotate-a-triangle proof for the angle-sum formulas.

Second Proof

Step 1. Suppose P = a+i b and Q = c+id are at unit distance from the origin of the plane. Then a2+b2 = 1 and c2+d2 = 1. (Units of length are omitted. The product P路Q has length 1.) The diagram below shows that the product P路Q located by adding angles (and multiplying lengths) coincides with the vector given by the sum of a路Q and i b路Q. This implies the distributive law (a+ib)路z = a路z+ib路z for the situation depicted. The argument holds regardless of the quadrants in which P is located.



The product a路Q is collinear with the vector Q as a is real while the product ib路Q is perpendicular to Q since ib has angle 90 degrees with the positive (real) axis.

Step 2. The foregoing implies the distributive law (a+ib)路z = a路z+ ib路z for case a2+b2 = 1 = |z|. Since the polar coordinate defined product of points in the plane is commutative, the foregoing law is two-sided. Hence z = c+id with c and d both real and c2+d2 = 1 implies

(a+ib)(c+id)
=
(a+ib)路z
=
a路z + ib路z by step 1
=
a路(c+id) + ib路(c+id)
=
a路c+a路id + b路ic+ib路id by step 1 twice
=
ac+iad+ibc+i2bd
=
ac-bd+i(ad+bc) as i2 = -1
due to the definition of the product of complex numbers and the associativity (proof?) of the addition of points in the plane. The real part of the product is ac-bd and the imaginary part is ad+bc.

Step 3. From the polar coordinate definition of the product of complex numbers, we observe

[cos(q)+isin(q)][cos(b)+isin(b)] = cos(q+b)+isin(q+b).
Now the angle sum formulas for sine and cosine:
cos(q)cos(b)-sin(q)sin(b) = cos(q+b)
and
cos(q)sin(b)+sin(q)cos(b) = sin(q+b)
follow from the expressions derived above for the rectangular coordinates of the product of two unit magnitude points in the plane. How to compute the real and imaginary parts in a product of any pair of complex numbers in terms of the real and imaginary parts of the factors is given next.

Distributive Law and Consequences

The next pages offer another way to obtain the distributive law for multiplication without assuming the factors have unit length. The next pages further imply the formulas for the real and imaginary parts of a product, and from them the angle sum formulas. This shows links between complex numbers and trigonometry can be obtained in different ways.
The previous sections show how multiplying points (a,b) in the plane with polar coordinate-based add the angle, multiple the lengths rule, led to the expression for the product in terms of the rectangular components, that is, the real and imaginary parts of the factors. This section provides an alternate approach. The expressions are consequences of the distributive law for complex multiplication.

The Commutative Law

Since the order of multiplication of positive numbers, and the order of addition of real numbers is immaterial, the add the angles, multiply the lengths rule implies

[r1,q1]路[r2,q2] = [r1r2,q1+q2] = [r2r1,q2+q1]
Therefore
[r1,q1]路[r2,q2] = [r2,q2]路[r1,q1]
This says that the order of multiplication is not important. Therefore, the right and left distributive laws imply each other (why?). So if one holds, then so does the others.

Use of The Distributive Laws

For every triple of complex numbers (A,B,C), the right distributive law (the right form)
A路{B+C} = A路B+A路C
and the left distributive law (the left form)
{B+C}路A = B路A+B路C
applied twice in succession would imply the following in terms of rectangular coordinates if they held.
(a,b)路(c,d)
=
(a,b)路{(c,0)+(0,d) }
=
(a,b)路(c,0)+(a,b)路(0,d)
if the right distributive law holds,
=
{(a,0)+(0,b)}路(c,0)+{(a,0)+(0,b)}路(0,d)
=
{(a,0)路(c,0)+(0,b)路(c,0)}
+{(a,0)路(0,d)+(0,b)路(0,d)}
if the left distributive law holds,
=
{(ac,0)+(0,bc)}
+{(0,ad)+ b[1,90掳]路d[1,90掳]}
=
{(ac,0)+(0,bc)}+{(0,ad)+ bd[1,180掳]}
=
{(ac,0)+(0,bc)}+{(0,ad)+ bd(-1,0)]}
=
{(ac,0)+(0,bc)}+{(0,ad)+ (-bd,0)}
=
{(ac,bc)}+{(-bd,ad)}
=
(ac-bd,bc+ad)
In terms of complex number notation, the foregoing says that
(a+ib)路(c+id) = (ac-bd)+(bc+ad)i
where i = 脰(-1). Therefore
(a+ib)路(c+id) = (ac-bd)+(bc+ad)i
holds for all real numbers a, b, c and d IF the left and right distributive laws hold.

Now

cis(a) = cos(a)+isin(a)
The property cis(a)路cis(b) = cis(a+b) follows from the add the angles, multiply the lengths definition of complex multiplication and not both factors have unit lengths. But this property cis(a)路cis(b) = cis(a+b) can be rewritten in terms of rectangular coordinates or complex number notation as
(cos(a),sin(a))路(cos(b),sin(b)) = (cos(a+b),sin(a+b))
Computation of the real and imaginary parts of the left hand side implies the angle-sum formulas for the cosine function
cos(a)cos(b)-sin(a)sin(b) = cos(a+b)
and for the sine function
cos(a)sin(b)-sin(a)cos(b) = sin(a+b)
respectively.

Proof of Distributive Laws

Plan. The proof of the distributive law A(P+Q) = AP+AQ will be based on the observation (the physical assumption) that multiplication by

A = [r,q] = [r,0]路[1,q] = [1,q]路[r,0]
can be done into two steps. One step is a rotation through the angle q while the other is a multiplication by the stretch factor or shrinkage factor r = [r,0]. Multiplication by a stretch factor and rotation through an angle will be shown to be distributive operations over addition.

Distributive Law For Stretch Factors. Now let P = (a,b) and Q = (c,d). Now

(r,0)*(P+Q)
=
(r,0)*[(a,b)+(c,d)]
=
(r,0)*(a+c,b+d)
=
(r{a+c},r{b+d})
=
(ra+rc,rb+rd)
=
(ra,rb)+(rc,rd)
=
(r,0)*P+(r,0)*Q
Therefore A(P+Q) = AP+AQ when A = (r,0) for some r > 0. This argument assumes the distributive law for multiplication of the sum of two real numbers by another.
It can be illustrated by tiling the plane with parallelograms - copies of the parallelogram determined by the arrows [1/(n)] P and [1/(n)]Q (where n 鲁 1 is a whole number). Such an illustration might be sufficient corroboration for some pre-algebraic students.

Distributive Law for Rotations. A parallelogram corresponding to the map addition of the arrows associated with P = (a,b) and Q = (c,d) is indicated below.

We assume that the parallelogram and the two triangle forming it are rigid bodies. This implies that after a rotation, that the map addition of the vectors forming the sides before and after rotation will yield the diagonal arrow before and after rotation, respectively. See the next diagram.

The following diagram shows that the triangle vertices P, Q and P+Q rotated respectively into the triangle vertices P垄, Q垄 and P垄+Q垄.

This suggests that

[1,q]路(P+Q) = P垄+Q垄 = [1,q]路P+[1,q]路Q
and hence that multiplication by the factor [1,q] is distributes over the addition of arrows.

End of the Proof. Observe that

A(P+Q)
=
([r,0]路[1,q])路(P+Q)
=
[r,0]路([1,q]路(P+Q))
=
[r,0]路([1,q]路P+[1,q]路Q)
=
[r,0]路([1,q]路P)+[r,0]路([1,q]路Q
=
([r,0]路[1,q])路P +([r,0]路[1,q])路Q
=
A路P +A路Q

Remark. The formal or proper presentation of mathematics relies on no diagrams and on no physical interpretation or reasoning. The preceding presentation of complex numbers was informal. It relied on geometric ideas (assumptions) to make a link between polar and rectangular coordinates. But the conclusions drawn above can be obtained in a geometric-free manner (no diagrams) and drawn solely from assumptions about arithmetic. See the diagram-free description of the complex numbers and trig functions in the university-level book Principles of Mathematical Analysis by W. Rudin, McGraw-Hill 1964, for more details.

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