New (August 3, 2001): Two webpages Complex Numbers and this one, Distributive Law for Complex Numbers. in original contributions to mathematics or mathematics education offer a short or shortest way to reach and explain trigonometry, the Pythagorean theorem, trig formulas for dot- and cross-products, the cosine law and a converse to the Pythagorean Theorem. Is there is a quicker way to learn or teach core ideas in trig and complex numbers? (After or besides the local material, visit chapters 1, 2 and 3 of Functional Trigonometry by Hillman and Alexanderson in the 1970's and posted online by Mervin E. Newton of Thiel College. It has slightly different starting point and more material.)
Step III. Distributive Law for Multiplication Over Addition
In the previous two steps, we saw how to add and multiply points in the plane using rectangular coordinates and complex number notation for those points. The previous steps implied this addition and multiplication of points satisfied all field axioms but one. This page derives the remaining field axiom, namely the distributive law for multiplication over addition. We will be using rectangular coordinates [a, b] during most of this derivation (a cosmetic choice) in place of writing a + bi. The derivation employs a few geometric assumptions.
In pure mathematics (real and complex analysis). it is possible to introduce to the complex numbers and trig functions too without drawing diagrams, that is, without employing geometric ideas and arguments except for illustration.. But that introduction or route is not for novices. See Principles of Mathematical Analysis by W. Rudin for the missing details if you are an advanced student.
Assumptions from Geometry in the Plane
- AAA Similarity Assumption (Postulate). Suppose corresponding angles in two triangles ABC and DEF are equal, then corresponding sides are proportional.
- SAS Congruency Assumption (Postulate). "Equality of two sides and the included angles for a pair of triangles" implies the triangles are isometric
- If in a quadrilateral ABCD, opposite sides have equal length then opposite sides are parallel and the quadrilateral is a parallelogram.
- Two distinct points in the plane uniquely determine a straight line.
- Given a line L and a point P not on the line, there is one and only one line M through P parallel to L.
- Two lines L and M meet in a single point or they coincide.
- Two nonparallel lines L and M uniquely determine a point P, their intersection point.
Addition of points in the plane
Recall the coordinate definition: The sum of two points with the rectangular coordinates [a,b] and [c,d] is given by [a+c,b+d]. We therefore write
[a,b] + [b,d] = [a+c,c+d]
Axioms for real numbers imply addition of points in the plane is associative and commutative.
Parallelogram Addition Method
Let O = [0,0], P = [a,b] = r1·cis(q1), Q = [c,d] = r2·cis(q2) and T = [a,b] + [b,d] = [a+c,c+d] = r3·cis(q3). Assume P, Q and T are nonzero. Without loss of generality, assume q1< q2 .
Rigid Motion Assumption: Assume that adding a point [g, h] to the three vertices of a triangle shifts the triangle into an isometric copy of itself, with corresponding sides parallel.
This assumption implies adding P= [a,b] to the triangle with vertices 0=[0.0], Q = [c,d] and U = [c,0] yields an isometric triangle with corresponding vertices [a,b] + 0 = P T= [a,b]+[c.d] = P+ Q, and U' = [a,b] + U = [a+c, b]. So the corresponding side OQ are PT are equal in length and parallel. Similarly the line segments OP and QT are equal in length and parallel by adding [c,d] to the triangle with vertices 0, P and V'=[a,0].
Therefore the coordinate-based addition method implies the quadrilateral OPTQ is a a parallelogram.
Conversely, let OPTQ be a parallelogram with P = [a,b] = r1·cis(q1), Q = [c,d] = r2·cis(q2) Our aim is to show that T must be P+ Q = [a,b] + [b,d] = [a+c,c+d]. Here we observe T is the intersection point of two lines, namely the unique line through P parallel to line segment OQ, and namely the unique line through Q parallel to line segment OQ. But the coordinate based addition method gives P+ Q = [a,b] + [b,d] = [a+c,c+d] as the intersection point of a line through P parallel to line segment OQ, and a line through Q parallel to line segment OQ. Therefore T = P +Q. So the construction or identification of a parallelogram OPTQ is a geometric way to compute or recognize the location of [a+c,c+d] =[a,b]+[c,d].
Parallelogram Construction
Given two points P and Q non-collinear with the origin O in the plane, draw the line segments OP and OQ. These lines segments are part of lines uniquely determined the points O and P together, and O and Q together. Now draw a line J parallel to OP through Q and draw another line K parallel to OQ through P. The lines J and K are not parallel. Therefore they intersect at a point T. By the previous argument, T = P + Q.
Remark: We could start with the parallelogram method to define the addition of points (or vectors) in the plane. Then the associative law for addition follows not by an appeal to the associative law for real numbers, but from the associativity of head-to-tail addition of vectors and a delicate discussion of equality for vectors before and during addition operations. See site area Complex No,s, Trig & Geometry for details.
Polar Coordinates and Multiplication by c > 0
Each point [a, b] in the plane has polar coordinates (r,q) where r = the distance of the point [a, b] in the plane and q is an angle, with say 0 < q < 360 degrees, uniquely determine by the rectangular coordinates a and b. So write
[a, b] = (r,q).
We assume by measurement, that polar coordinates can be obtained from rectangular coordinates, and vice-versa.
Given c > 0, put
c·[a,b] = (ca,cb)
Similarity of right triangles in the first to fourth quadrants (4 cases/diagrams)
c·[a,b] = (cr,q)
when a and b are both nonzero. The cases where one or both are zero
are left to the reader.
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Algebraic Argument: Suppose P = [x1, y1], Q= [x2, y2]. Then
This chain of reason shows algebraically how multiplication by c > 0 distributes over addition. In it, the identity
was justified in part by a geometric appeal to the similarity of right triangles. |
Polar Coordinates and the Function cis(q)
Let
cis(q) = (1,q).
be the point on the unit circle with angle q. Then
[a, b] = (r,q) = r · (1,q) = r · cis(q)
Unit Circle Definition of sine, cosine and Tangent functions
In terms of complex number notation, we may let the cosine and sine functions of be defined as follows
- cos(q) = Re( cis(q)) = the real part of cis(q), and
- sin(q) = Im( cis(q)) = the imaginary part of cis(q)
so that
cos(q) + i sin(q) = cis(q) = [cos(q), sin(q) = (1,q)
is a point on the unit circle. The tangent of q is given by
tan(q) = cos (q) / sin(q)
when sin( ) is nonzero.
Ratio of Sides Computation of sines and cosines.
For 0 < q < 90 degrees, similarity of right triangles gives the usual ratios
cos q = adjacent / hypotenuse
sin q = opposite / hypotenuse
as an alternative way to compute cosines and sines. Trigonometric courses should give the unit definition of sine and cosine first, or give an explanation of how the similarity of right triangles implies the usual ratios depend only the angle q.
Multiplication of Points in the Plane
In polar coordinate notation, the product of two points in the plane is given by
(r1,q1)·(r2,q2) = (r1r2,q1+q2) or{r1·cis(q1) } · {(r2 · cis(q2)} = (r1r2) · cis(q1+q2)
Axioms for real numbers immediately imply this multiplication is commutative and associative.
Distributive Law for Multiplication over Addition
The associativity of multiplication, and the identity
[A, B] = (r,q) = r · (1,q) = r · cis(q)
implies the full distributive law for multiplication over addition follows if we establish it for
- multiplication by r = c > 0 and
- by the unit circle point cis(q) = (1,q)
Details follow for both cases, albeit case I was shown above. There is some repetition here.
Case I. Multiplication by r = c > 0
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Algebraic Argument (Repeated): Suppose P = [x1, y1], Q= [x2, y2]. Then
This shows algebraically how multiplication by c > 0 distributes over addition, albeit above the identity c·[x1, y1] = [cx1,cy1] was justified by an appeal to the similarity of right triangles. |
Geometric Argument: Let P = (r1,q1) and Q= (r2,q2) be points in the plane, non-collinear with the origin. Without loss of generality, we may assume q1 < q2. Swap the points P and Q if need-be. Let T = P + Q = (r3,q3) be their sum.
Then P, Q and T are vertices of a parallelogram OPTQ.
Multiply T = P + Q = (r3,q3) by c > 0 to obtain a point T' = (cr3,q3) = c · T on the line through OT. See diagram.
In the diagram, the line through T' parallel to OP intersects the line through OQ at a point Q'. Similarly, the line through T' parallel to OQ intersects the line through OQ at a point P'. From the similarity of triangles OPT and OP'T', P' = (cr1,q1) = c · P. Likewise, from the similarity of the triangles OQT and OQ'T', Q' = (cr2,q2) = c·Q. Therefore
c·(P+Q) = c·P +c·Q.
This shows that multiplication by c > 0 distributes over addition at least when O, P and Q are non-collinear. The collinear case follows by a discussion of continuity.
Case II. Multiplication by R= cis(q) = (1,q)
Let P = (r1,q1) and Q= (r2,q2) be points in the plane, non-collinear with the origin. Without loss of generality, we may assume q1 < q2. Swap the points P and Q if need-be. Let T = P + Q = (r3,q3) be their sum.
Then P, Q and T are vertices of a parallelogram OPTQ.
Multiplication by R = cis(q)= (1,q) yields points P'(r1,q1+q), Q'= (r2,q2 + q) and T' = (r3,q3 + q). Here the SAS (side angle side) isometry postulate implies the triangles POT and P'OT' are congruent. Similarly, triangles QOT and Q'OT' are congruent. Congruencies together imply OP'T'Q' is a quadrilateral in which opposite sides have equal length. Therefore this quadrilateral is a parallelogram.
In consequence
T' = P' + Q'
by the parallelogram addition method. The latter may be rewritten as
R (P+Q) = RP + RQ
Thus multiplication by R= cis(q) = (1,q), a rotation through the angle q, distributes over addition when O, Q and P are not collinear.
The collinear two collinear cases ( OP and OQ in the same or opposite direction) and the case where one or both points P and Q are at the origin is left to the reader.
Proof of Distributive Law.
Let Z = (r , q) = r . cis (q) with r > 0. Suppose P and Q are points in the plane.
Therefore
Z(P+Q) = (r . cis (q)(P+Q)
= r . ( cis (q)(P +Q) ) by associativity
of multiplication.= r . ( cis (q)P + cis (q)Q ) as rotation distributes
(case II above)= r . (cis (q)P) + r . (cis (q)Q ) as scalar multiplication
distributes (case I)= (r . cis (q)P + (r . cis (q)Q by associativity
= ZP +ZQ
This implies the right distributive law
Z(P+Q) = ZP +ZQ
The left distributive law
(P+Q) Z = PZ + QZ
follow from the right by commutivity of multiplication.
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Complex Numbers |
The fundamental theorem of algebra and partial fraction decomposition in calculus depend on complex numbers.
Easy Consequences
Hint: See the (newest) Complex
Number. Starter Lesson.
for a simple geometric introduction, then continue with easy consequence below.
The clearest geometric proof of the distributive law appears in the
folder.
First Earlier (Old) exposition of complex numbers follows in Z1 to B1 below - read for review or revision .
D1 to D6 after provide a review of vectors.
More on Complex Numbers:
Chapters in Volume 3::
Intro assumes the field properties
of real numbers in place of
deriving them geometrically