Introduction to the Geometry of Complex Numbers

By Roland Deaux

Geared toward readers unfamiliar with complex numbers, this text explains how to solve the kinds of problems that frequently arise in the applied sciences, especially electrical studies. To assure an easy and complete understanding, it develops topics from the beginning, with emphasis on constructions related to algebraic operations.
The three-part treatment begins with geometric representations of complex numbers and proceeds to an in-depth survey of elements of analytic geometry. Readers are assured of a variety of perspectives, which include references to algebra, to the classical notions of analytic geometry, to modern plane geometry, and to results furnished by kinematics. The third chapter, on circular transformations, revives in a slightly modified form the essentials of the projective geometry of real binary forms. Numerous exercises appear throughout the text.

• Language: English
• Publisher: Dover Publications
• Release date: Jan 23, 2013
• ISBN: 9780486158044

Book preview

(E)

CHAPTER ONE

GEOMETRIC REPRESENTATION OF COMPLEX NUMBERS

I. FUNDAMENTAL OPERATIONS

1. Complex coordinate. Consider the complex number

x + iy    (x and y real)

which we denote by z. Draw, in a plane, two perpendicular coordinate axes Ox, Oy. The point Z having for abscissa the real part x of the number z and for ordinate the coefficient y of i is called the representative point, or the image, of the number z. Conversely, each real point Z1, of the plane is the image of a unique complex number z1 equal to the abscissa of Z1 increased by the product with i of the ordinate of this point. The number z1 is called the complex coordinate, or the affix, of the point Z1.

A plane in which each real point is considered as the image of a complex number is called the Gauss plane, the Cauchy plane, or the plane of the complex variable.

We shall denote a point of the Gauss plane by an upper case letter, and its affix by the corresponding lower case letter.

Corollaries. 1° The Ox axis is the locus of the images of the real numbers. The Oy axis is the locus of the images of the pure imaginary numbers. This is why Ox and Oy are sometimes called the real axis and the imaginary axis of the Gauss plane.

2° The number—z is the affix of the symmetric of point Z with respect to the origin O.

2. Conjugate coordinates. The complex number conjugate to

z = x + iy

will be designated by the notation

z = x – iy

which is read, " z .

The image of the number is the point symmetric to Z with respect to the Ox axis.

3. Exponential form. In the Gauss plane, we choose for the positive sense of rotation and of angles the sense of the smallest rotation about O which carries the Ox axis into the Oy axis. The algebraic value (xy) of any angle having Ox for initial side and Oy for terminal side is, then, to within an integral multiple of 2π,

(1)

Let Z be the image of a non-zero complex number

(2)

FIG. 1

Place an arbitrary axis a on the line OZ.

Denote by θ and r the algebraic values of any one of the angles (xa) and of the segment OZ. One recognizes in θ and r polar coordinates of Z for the pole O and the axis Ox. From the theorem on orthogonal projections we have

x = r cos θ

and, because of equation (1),

y = r sin θ.

Equation (2) can then be written as

z = r (cos θ + i sin θ

or

(3)

the exponential form of z.

When z = 0, we are to take r zero and θ arbitrary.

We have

4. Case where r is positive. When the positive sense of the a axis is that from O toward Z, the number r is positive and is the modulus of z; θ is then an argument of z. We write

the radical signifying that we extract the arithmetic square root of x² + y².

If, on the contrary, the a axis is such that r is negative, then equation (3) can be written as

z = ( – r)( – 1) eiθ

or, since – 1 = eiπ, as

z = ( – r) ei(π+θ).

The modulus and an argument of z are then – r and π + θ.

5. Vector and complex number. The image Z of the number z , the vectorial coordinate of Z for the pole O. We can then say that the number z is represented by this vector. The number and the vector have equal moduli, and we can conveniently speak of an argument of the number as an argument of the vector.

Nevertheless, we will never convey these facts by writing

as is done by some authors, for such a use of the = sign easily leads to contradictions when employed in connection with the product of two vectors (see article 10) in the sense of classical vector analysis.

6. Addition. If the n complex numbers

zk = xk + iyk   (k = 1, 2,... , n)

have for images the n points Zk, their sum

(4)

has for image the point Z defined by the geometric equation

(5)

FIG. 2

Let x, y be the coordinates of the point Z constructed with the aid of equation (5). By algebraically projecting first on the Ox axis, then on the Oy axis, we obtain the two algebraic equations

x = ∑Xxk,   y = ∑yk

and hence

x + iy = ∑ (xk + iyk),

that is to say, equation (4).

7. Subtraction. If the complex numbers z1 z2 are represented by the vectors the difference

z = z1 – z2

is represented by the geometric difference

of the corresponding vectors.

We have

is the symmetric of Z2 with respect to O.

FIG. 3

By virtue of article 6 we have

(6)

Corollaries. 1° Any vector of the Gauss plane represents the complex number equal to the complex coordinate of the extremity of the vector diminished by the complex coordinate of the origin of the vector.

Equation (6) gives, in effect,

, the difference z1 – z2.

2° Any equality between two geometric polynomials whose terms are vectors of the Gauss plane is equivalent to an equality between two algebraic polynomials whose terms are the complex numbers represented by these vectors, and conversely.

Thus, using the notation suggested in article 1, an equation such as

is equivalent to the algebraic equation

b – a = d – c + f – e.

8. Multiplication. If the complex numbers z1, z2 are represented by the vectors , the product

z = z1 z2

is represented by the vector which one obtains from , for example, as follows : 1° rotate , about O through an angle equal to the argument of the other vector ; 2° multiply the vector thus obtained by the modulus of vector .

If r1, r2 and θ1, θ2 are the moduli and the arguments of z1, z2, we have, by (3) of article 3,

Therefore

The argument of z is then θ1 + θ2, while its modulus is

which justifies the indicated construction.

FIG. 4

We can realize the same end by taking on Ox the point U having abscissa + 1. The sought point Z is the third vertex of triangle OZ1Z directly similar to triangle OUZ2, for

Particular cases. 1° The number z2 is real. Its argument is 0 or π according as it is positive or negative.

We have

and the point Z is on the line OZ1.

2° The number z2 is complex with unit modulus. It then has the form

and is obtained from Z1 by a rotation of angle θ2 about O.

3° The number z2 is i or – i. Since

i = eiπ/²,   – i = e – iπ/²,

to multiply a complex number z1 by ± i is to rotate its representative vector , about O through an angle of ± π/2.

9. Division. If the complex numbers z1, z2 are represented by the vectors , the quotient

is represented by the vector which one obtains from the vector as follows : 1° rotate about O through an angle equal to the negative of the argument of vector ; 2° divide the vector thus obtained by the modulus of vector .

Using the notation of article 8, the construction of Z follows from

The point Z is the third vertex of triangle OZ1Z directly similar to triangle OZ2U.

FIG. 5

We thus treat division as the inverse operation of multiplication.

Particular case. Construction of the point Z given by z = 1/z2. Since z1 = 1, the point Z1 is at U. The lines OZ2, OZ are symmetric with respect to Ox and we have

is the symmetric of Z with respect to Oxcorrespond to one another under the inversion having center O and power 1; that is to say, they are harmonic conjugates with respect to the extremities P, Q of the diameter of the circle having center O and radius OU.

FIG. 6

Consequently, the point Z is the Symmetric with respect to Ox of the inverse of point Z2 under the inversion having center O and power 1.

Corollaries. 1° If a point A has affix a, then its inverse under the inversion having center O and power 1 has affix 1/ā and not 1/a.

2° The construction of the point with affix

can be reduced to that of article 8 by writing

10. Scalar product of two vectors. If z1, z2 are the affixes of the points Z1, Z2, then

In fact, from the relations

we find

and, in the same way,

It suffices to replace x1, y1, x2, y2 by these values in the classical expression

x1x2 + y1y2

in order to obtain the announced result.

Corollary. If a is the affix of the point A, and if , are two vectors in the Gauss plane, then (7, corollary 1°)

11. Vector product of two vectors. If z1, z2 are the affixes of the points Z1, Z2, the algebraic value of on an axis Oζ such that the trihedral Oxyζ is trirectangular and right-handed, is equal to

This algebraic value is, in fact,

x1y2 – x2y1

and it suffices to replace x1, x2, y1, y2 by the values given in article 10 in order to obtain the announced result.

Corollaries. 1° The algebraic value considered is twice that of the area of triangle OZ1Z2.

2° If a is the affix of point A, and if are two vectors of the Gauss plane, we have (7, corollary 1°) for the algebraic value of on an axis Oζ such that the trihedral Oxyζ is trirectangular and right-handed

12. Object of the present course. To each complex number z corresponds a point Z, and conversely. To each of the fundamental operations performed on complex numbers corresponds a geometric construction (6-9). Consequently, to each algebraic operation performed on complex numbers z1, z2, ..., zn, and to each property of such an operation, corresponds a construction concerning the points Z1, Z2, ..., Zn and a property of the figure obtained.

To interpret this passage from the algebraic manipulation of complex numbers to the geometric concept, or the reverse, is the aim of the present notes.

Exercises 1 through 11

1. Construct the sets of points having for affixes :

2. Construct the images of the roots of the following equations:

z² – 1 = 0,  z² + 1 = 0,  z³ = 0,  z³ + 1 = 0,  z⁴ – 2 = 0,  z⁴ + 1 = 0,   z⁶ – 1 = 0  z⁶ + 1 = 0,  z⁸ – 1 = 0,  z⁸ + 1 = 0.

If n is a positive integer and a a number with modulus r and argument θ, what can be said about the figure formed by the images of the roots of the equations

zn – a = 0,    zn + a = 0?

3. Which of the numbers represented by the following expressions are real and which are pure imaginary?

4. If a is the affix of a point A, construct the points with affixes:

[Employ the exponential form in the last two cases.]

5. Distance between two points. The distance between two points A, B is

Using the identity

x² + y² = (x + iy) (x – iy)

and the elements of analytic geometry, show that:

;

2° the square of the distance between points A and B is

6. Angles between two axes. Remembering that if two axes p, q we have, where (p,q, show that:

is given by

where the radicals denote the arithmetic square roots;

2° the angle θ and the Ox axis is given by

expressions which are more easily obtained from z = | z | eiθ.

7. Equilateral triangle. A necessary and sufficient condition for three points A, B, C with affixes a, b, c to be vertices of an equilateral triangle is that

a relation which is equivalent to any one of the following:

or, again, that b – c, c – a, a – b be roots of an equation of the form

z³ – k = 0.

[Set

b – c = α,   c – a = β,   a – b = γ,

whence

α + β + γ = 0.

It is necessary that

relations which, with

give

Conversely, if

α² = βγ,

whence

then

See article 84 for another demonstration.]

8. With the aid of exercise 7, show that the images of the roots of the cubic equation

z³ + 3a1z² + 3a2z + a3 = 0

form an equilateral triangle if a1² = a2. Deduce that the origin and the images of the roots of the equation

z² + pz + q = 0

form an equilateral triangle if p² = 3q.

9. The lines OZ1, OZ2 are perpendicular or parallel according as

The lines Z1Z2, Z3Z4 are perpendicular or parallel according as

or

[Use articles 10 and 11.]

10. If ab = cd, we have

| OA | | OB| = |OC | | OD |

have the same bisectors. [Use moduli and arguments.]

If ab = c. What can be said if ab = – c²?

11. If a, b, c are numbers of modulus 1 and if we set

s1 = a + b + c,   s2 = ab + bc + ca,   s3 = abc,

show, by employing

that

, .

II. FUNDAMENTAL TRANSFORMATIONS

13. Transformation.

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