Complex numbers

Topic M–11–2

Complex Numbers

Contents 1 Basic Definitions 1.1 Representation of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . .

2 2

2 Operations on complex numbers 2.1 Basic operations . . . . . . . . . . . . 2.2 Powers and roots of complex numbers 2.2.1 Powers . . . . . . . . . . . . . 2.2.2 Roots . . . . . . . . . . . . . 2.2.3 Solution of a binomial equation

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4 4 7 7 8 9

multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10 10 11 11

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13

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13 14

3 Geometry of complex numbers 3.1 Euler’s formula. Rotation as complex 3.2 Conjugate . . . . . . . . . . . . . . 3.3 Modulus — the distance µ function ¶ . . z1 − z0 3.4 Interpretation of arg . . z2 − z0 3.5 Lines in the complex plane . . . . . 3.6 Circles in the complex plane . . . . .

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2

Basic Definitions

1

Basic Definitions

¤ Definition 1 (Complex Number). A complex number is a number given by the expression z = a + i b,

(1)

where a, b ∈ R, the set of real numbers, and i is the so–called imaginary unit, which is defined as √ i = −1 or i2 = −1. (2) The set of all complex numbers is denoted as C. Further, a is called the real part, and b, the imaginary part of the complex number z. They are denoted, respectively, as follows: a = Re (z),

b = Im (z).

If Re (z) = 0, the complex number z is said to be purely imaginary, while if Im (z) = 0, z is purely real. Thus, all real numbers are also complex, i.e. R ⊂ C. Two complex numbers z = a + i b and z = a − i b that differ solely in the sign of the imaginary part are called conjugate complex numbers. We agree upon the two following basic definitions: 1. Two complex numbers z1 = a1 + i b1 and z2 = a2 + i b2 are equal, z1 = z2 , if and only if a1 = a2 ,

and

b1 = b2 ,

that is, if their real parts are equal and their imaginary parts are equal. 2. A complex number z is equal to zero, z = a + ib = 0 if and only if a = 0, b = 0.

1.1

Representation of complex numbers

1. Geometric representation of complex numbers. Any complex number z = a + i b may be represented in the xy plane as a point P (a, b) with coordinates a and b. Conversely, every point M (x, y) of the plane is associated with a unique complex number z = x + i y. Thus, there is a one–to–one correspondence between the set of complex numbers, C, and the xy plane. This is symbolically shown in Figure 1. The plane on which the complex numbers are represented is called the plane of the complex variable, z, or the complex plane, or the Argand plane (Figure 1 the encircled z symbol indicates that this plane is the complex plane). Points of the complex plane lying on the x–axis correspond to real numbers (Im (z) = 0). Points on the y–axis represent purely imaginary numbers, since Re (z) = 0. Therefore, in the complex plane, the y axis is called the imaginary axis, and the x axis is the real axis. −→ Joining the point P (a, b) to the origin O, we get a vector OA. In certain instances, it is −→ convenient to treat the vector OA itself as the geometric representation of the complex number z = a + i b.

Anant Kumar

• study circle for iitjee & aieee •

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Basic Definitions y

z = a + ib

P (a, b)

P (a, b)

C

z

y

z

O

r

x

b

ϕ a

O

Set of complex numbers

x

Figure 1: The correspondence between the set of complex numbers and

Figure 2: The polar and the vec-

xy plane

tor representation of the complex number z = a + i b.

2. Trigonometric form of a complex number. Denote by ϕ and r (where r ≥ 0) the polar coordinates of the point P (a, b) and consider the origin as the pole and the positive direction of the x axis, the polar axis. Then (Figure 2) we have the familiar relationships a = r cos ϕ,

b = r sin ϕ

and, hence, the complex number may be written in the form z = a + i b = r cos ϕ + i r sin ϕ

or

z = r(cos ϕ + i sin ϕ).

(3)

The expression on the right is called the trigonometric form (or polar form) of the complex number z = a + i b. r is termed the modulus of the complex number z, while ϕ is the argument (or phase) of the complex number z. They are designated as r = |z|,

ϕ = arg(z).

The quantities r and ϕ are expressed in terms of a and b as follows: r=

p

a2 + b2 ,

ϕ = tan−1

b . a

To summarize, then,

 √ a2 + b2  (4) b  arg(z) = arg(a + i b) = tan−1 a The argument of a complex number is considered positive if it measured from the positive x axis in the counterclockwise, and negative, in the opposite sense. The argument ϕ is obviously not determined uniquely but up to term 2kπ where k is an integer. |z| = |a + i b| =

♣ Remark 1. The value of ϕ satisfying the inequalities −π < ϕ ≤ π is termed as the principal argument. ♣ Remark 2. The conjugate complex numbers z = a + i b and z = a − i b have equal moduli |z| = |z| and their arguments are equal in magnitude but differ in sign: arg(z) = − arg(z). ♣ Remark 3. Any real number A can also be written in the polar form, namely: ½ |A| (cos 0 + i sin 0) for A > 0 A= |A| (cos π + i sin π) for A < 0

(5)

The modulus of the complex number 0 is zero: |0| = 0. Any angle ϕ may be taken for the argument of 0. Indeed for any angle ϕ, we have 0 = 0 (cos ϕ + i sin ϕ). Anant Kumar


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Operations on complex numbers

2 2.1

Operations on complex numbers Basic operations

1. Sum of complex numbers. The sum of two complex numbers z1 = a1 +i b1 and z2 = a2 +i b2 is a complex number defined by the equation z1 + z2 = (a1 + i b1 ) + (a2 + i b2 ) = (a1 + a2 ) + i (b1 + b2 ).

(6)

From (6), it follows that the addition of complex numbers depicted as vectors is performed by the rule of addition vectors (see Figure 3(a)). Im

Im

z1 z1 z1

+

z2

z2

z1 − z2

z2 Re

O

Re

O

(a)

(b)

Figure 3: The addition and subtraction of complex numbers follow the same rules as those of vectors. 2. Subtraction of complex numbers. The difference of two complex numbers z1 = a1 + i b1 and z2 = a2 + i b2 is a complex number such that when it is added to z2 , it yields z1 . It is easy to see that z1 − z2 = (a1 + i b1 ) − (a2 + i b2 ) = (a1 − a2 ) + i (b1 − b2 ).

(7)

It will be noted that the difference of two complex numbers z1 and z2 is represented by the displacement vector from the point on the complex plane represented by z2 to that represented by z1 (see Figure 3(b)). Thus, the modulus of z1 − z2 is the distance between the two points: p |z1 − z2 | = (a1 − a2 )2 + (b1 − b2 )2 . 3. Multiplication of complex numbers. The product of two complex numbers z1 = a1 + i b1 and z2 = a2 + i b2 is a complex number obtained when these two numbers are multiplied as binomials by the rules of algebra, provided that i 2 = −1,

i 3 = −i,

i 4 = (i 2 )2 = (−1)2 = 1,

i5 = i,

etc.

and, generally, for any integral k, i 4k = 1,

i 4k+1 = i ,

i 4k+2 = −1,

i 4k+3 = −i

From this rule we get z1 z2 = (a1 + i b1 )(a2 + i b2 ) = a1 a2 + i b1 a2 + i a1 b2 + i 2 b1 b2 ⇒

z1 z2 = (a1 a2 − b1 b2 ) + i (b1 a2 + a1 b2 )

(8)

Let the complex numbers be written in trigonometric form z1 = r1 (cos ϕ1 + i sin ϕ1 ),

Anant Kumar

z2 = r2 (cos ϕ2 + i sin ϕ2 ).


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Operations on complex numbers Im

Im

r2

P

r1

r2 P2 (z2 )

ϕ1 ϕ2

O

P2 (z2 )

r2

ϕ2

P1 (z1 )

O

r1

E

Re

ϕ1

P1 (z1 )

r1

ϕ1

ϕ2

Re

E

P (a)

(b)

Figure 4: Finding out the product and quotient of complex numbers using similar triangles. Then using the expression (8), we obtain z1 z2 = r1 r2 [(cos ϕ1 cos ϕ2 − sin ϕ1 sin ϕ2 ) + i (sin ϕ1 cos ϕ2 + cos ϕ1 sin ϕ2 )] = r1 r2 [cos(ϕ1 + ϕ2 ) + i sin(ϕ1 + ϕ2 )]

(9)

that is, the product of two complex numbers is a complex number, the modulus of which is equal to the product of the moduli of the factors, and the argument is equal to the sum of the arguments of the factors. Let us derive a geometrical interpretation of the product of two complex numbers. Let P1 be the point representing the complex number z1 while P2 correspond to z2 (Figure 4(a)). Let E be a point on the real axis such that OE = 1 unit. Complete the triangle OP1 E. Now taking OP2 as the base, construct a triangle OP P2 similar to 4 OP1 E so that OP : OP1 = OP2 : OE. But, since OE = 1, so we get OP = OP1 · OP2 . Also, ∠P2 OP = ∠EOP1 = ϕ1 . Thus, ∠EOP = ϕ1 + ϕ2 . Hence, P represents the complex number for which the modulus is r1 r2 , and the argument is ϕ1 + ϕ2 . Thus, it represents the complex number z1 z2 . ♣ Remark 4. The product of two conjugate complex numbers z = a + i b and z = a − i b is, by virtue of (8), expressed as follows: z z = a2 + b2

⇒

z z = |z|2 = |z|2 .

(10)

4. Division of complex numbers. The division of complex numbers is defined as the inverse operation of multiplication. p z1 Suppose, we have z1 = a1 + i b1 , z2 = a2 + i b2 , |z2 | = a22 + b2 6= 0. Then, = z is a z2 complex number such that z1 = z2 z. If a1 + i b1 = x + i y, a2 + i b2 then a1 + i b1 = (a2 + i b2 )(x + i y)

⇒

a1 + i b1 = (a2 x − b2 y) + i (a2 y + b2 x).

x and y are found from the system of equations a1 = a2 x − b2 y, Anant Kumar

b1 = b2 x + a2 y.


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Solving this system of equations, we obtain x=

a1 a2 + b1 b2 , a22 + b22

y=

a2 b1 − a1 b2 , a22 + b22

and finally, we have the quotient z=

a1 a2 + b1 b2 a2 b1 − a1 b2 +i 2 2 a2 + b2 a22 + b22

(11)

Actually, complex numbers are divided as follows: to divide z1 = a1 + i b1 by z2 = a2 + i b2 , multiply the dividend and divisor by the conjugate of the divisor (that is, by a2 − i b2 ). Then the divisor will be a real number; dividing the real and imaginary parts of dividend by it, we get the quotient a1 + i b1 (a1 + i b1 )(a2 − i b2 ) (a1 a2 + b1 b2 ) + i (a2 b1 − a1 b2 ) = = a2 + i b2 (a2 + i b2 )(a2 − i b2 ) a22 + b22 a1 a2 + b1 b2 a2 b1 − a1 b2 = +i 2 2 a2 + b2 a22 + b22 If the complex numbers are in trigonometric form z1 = r1 (cos ϕ1 + i sin ϕ1 ),

z2 = r2 (cos ϕ2 + i sin ϕ2 )

then, using (11), we can obtain z1 r1 (cos ϕ1 + i sin ϕ1 ) r1 = = [cos(ϕ1 − ϕ2 ) + i sin(ϕ1 − ϕ2 )] z2 r2 (cos ϕ2 + i sin ϕ2 ) r2

(12)

Thus, the modulus of the quotient of two complex numbers is equal to the quotient of the moduli of the dividend and the divisor; the argument of the quotient is the difference between the arguments of the divisor and divisor. We shall use this result to obtain a geometric interpretation of the division of two complex numbers. As shown in Figure 4(b), let P1 and P2 be the affixes of z1 and z2 respectively. On OP1 construct the triangle OP P1 similar to 4OEP2 , where E lies on the real axis and OE = 1. Now r1 OP : OE = r1 : r2 ⇒ OP = r2 Also, ∠XOP = ϕ1 − ϕ2 . The point P thus represents the quotient z1 /z2 , since its modulus is r1 /r2 and its argument is ϕ1 − ϕ2 . ♣ Remark 5. From the rules of operations involving complex numbers it follows that the operations of addition, subtraction, multiplication and division of complex numbers yield a complex number. Also it is easy to show that if each complex number in these expressions is replaced by its conjugate, then the results of the aforementioned operations will yield conjugate numbers, whence, as a particular case, we have the following theorem. ¯ Theorem 1. If in a polynomial with real coefficients f (x) = an xn + an−1 xn−1 + · · · + a0 we put, in place of x, the number z = a + i b and then its conjugate z = a − i b, the results of the two operations will be mutually conjugate, that is f (z) = f (z)

Anant Kumar


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2.2 2.2.1

Powers and roots of complex numbers Powers

If in the expression (9), we substitute z1 = z2 = z = r(cos ϕ + i sin ϕ), we obtain z 2 = r2 (cos 2ϕ + i sin 2ϕ) Further, z 3 = z 2 z = r3 (cos 3ϕ + i sin 3ϕ). It can be proved by the principle of mathematical induction that for a positive integer n, [r(cos ϕ + i sin ϕ)]n = rn (cos nϕ + i sin nϕ)

(13)

This is called De Moivre’s formula. It shows that when a complex number is raised to a positive integral power, the modulus is raised to this power, and the argument is multiplied by the exponent. The above formula holds true for negative integers as well. For if n < 0, let n = −m for a positive integer m. Then, for the complex number z = r(cos ϕ + i sin ϕ), we proceed as follows: 1 zm 1 [r(cos ϕ + i sin ϕ)]m 1 (using De Moivre’s formula for positive integer) m r (cos mϕ + i sin mϕ) 1 cos mϕ − i sin mϕ r−m · cos mϕ + i sin mϕ cos mϕ − i sin mϕ cos mϕ − i sin mϕ r−m cos2 mϕ + sin2 mϕ r−m [cos(−mϕ) + i sin(−mϕ)]

z n = z −m = = = = = =

= rn (cos nϕ + i sin nϕ)

(since −m = n)

The validity of De Moivre’s formula in case of rational n will be taken up once we have discussed the roots of a complex number. For the moment, consider another application of De Moivre’s formula. Setting r = 1, in the formula (13), we get (cos ϕ + i sin ϕ)n = cos nϕ + i sin nϕ Expanding the left hand side by the binomial theorem and equating the real and imaginary parts, we can express sin nϕ and cos nϕ in terms of powers of sin ϕ and cos ϕ: µ ¶ µ ¶ µ ¶ n n n n n−2 2 cos nϕ = cos ϕ − cos ϕ sin ϕ + cosn−4 ϕ sin4 ϕ − . . . (14) 0 2 4 µ ¶ µ ¶ µ ¶ n n n n−1 n−3 3 sin nϕ = cos ϕ sin ϕ − cos ϕ sin ϕ + cosn−5 ϕ sin5 ϕ − . . . (15) 1 3 5 µ ¶ n Here, represents the binomial coefficients. For instance, if n = 3, we have k cos3 ϕ + i 3 cos2 ϕ sin ϕ − 3 cos ϕ sin2 ϕ − i sin3 ϕ = cos 3ϕ + i sin 3ϕ Making use of the condition for equality of two complex numbers, we get cos 3ϕ = cos3 ϕ − 3 cos ϕ sin2 ϕ sin 3ϕ = − sin3 ϕ + 3 cos2 ϕ sin ϕ

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2.2.2

Roots

The nth root of a complex number is another complex number whose nth power is equal to the radicand, or p n r(cos ϕ + i sin ϕ) = ρ(cos ψ + i sin ψ) ρn (cos nψ + i sin nψ) = r(cos ϕ + i sin ϕ)

if

Since the moduli of equal complex numbers must be equal, while their arguments may differ by an integral multiple of 2π, we have ρn = r,

and

nψ = ϕ + 2kπ

where k = 0, ±1, ±2, . . .

Whence we find ρ= Therefore, we obtain

√ n r,

and

ψ=

ϕ + 2kπ n

k = 0, ±1, ±2, . . .

µ ¶ √ ϕ + 2kπ ϕ + 2kπ n r(cos ϕ + i sin ϕ) = r cos + i sin n n

p n

(16)

√ Here n r is the principal (positive real) root of the positive number r. Giving k the values 0, 1, 2, . . . , n − 1, we get n different values of the root. For other values of k, the arguments will differ from those already obtained by a multiple of 2π, and, for this reason, root values will be obtained that coincide with those already obtained. Thus, the nth root of a complex number has n distinct values. The nth root of a real nonzero number real number A also has n values, since a real number is a special case of a complex number and may be trigonometric form: if A > 0, then A = |A| (cos 0 + i sin 0) if A < 0, then A = |A| (cos π + i sin π) Example 1. Find all the values of the cube root of unity. Solution: We represent unity in trigonometric form: 1 = cos 0 + i sin 0 Thus, using (16) we have √ 3

1=

√ 3

cos 0 + i sin 0 = cos

0 + 2kπ 0 + 2kπ + i sin , 3 3

k = 0, 1, 2

Setting the value of k equal to 0, 1, 2, we find three values of the root: ω

x0 = cos 0 + i sin 0 = 1,

Im

√ 2π 1 3 2π 120◦ + i sin =− +i , Re 3 3 2 O 1 √2 4π 4π 1 3 x2 = cos + i sin =− −i 3 3 2 2 ω2 √ 1 3 Usually, the number − + i is denoted as ω. It is then simple enough to show that 2 2 √ 1 3 2 ω =− −i 2 2 x1 = cos

Thus the three cube roots of unity are 1, ω, ω 2 . Note that 1 + ω + ω 2 = 0,

and

ω3 = 1

These three roots are geometrically represented as in the adjoining figure. Anant Kumar


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Getting back to De Moivre’s theorem, we can now see what will happen if n is a rational p number. Let n = where p and q > 0 are integers having no common factors. Then, q (cos ϕ + i sin ϕ)n = (cos ϕ + i sin ϕ)p/q = (cos pϕ + i sin pϕ)1/q (De Moivre’s formula for a positive integer) pϕ + 2kπ pϕ + 2kπ + i sin k = 0, 1, 2, . . . , q − 1 = cos q q Thus, when n is rational of the form p/q, there are q distinct values for (cos ϕ + i sin ϕ)n , one of which is cos nϕ + i sin nϕ. 2.2.3

Solution of a binomial equation

An equation of the form z n = x0

or

z n − x0 = 0,

where x0 is a constant, is called a binomial equation. Let us find its roots. If x0 is a real positive number, then the roots are given by ¶ µ √ 2kπ 2kπ n + i sin , k = 0, 1, 2, . . . , n − 1 zk = x0 cos n n The expression in the brackets gives all the values of the nth roots of unity. If x0 is real negative number, then the roots are given by µ ¶ p π + 2kπ π + 2kπ n zk = |x0 | cos + i sin , k = 0, 1, 2, . . . , n − 1 n n The expression in the brackets gives all the values of the nth roots of −1. If x0 is a complex number, then the values of zk are found from (16). Once we obtain the values of zk , we can express the binomial f (z) = z n − x0 as the product of n linear factors: z n − x0 ≡ (z − z0 )(z − z1 )(z − z2 ) · · · (z − zn−1 )

(17)

The nth roots of unity. As an illustration of the solution of a binomial equation, let’s solve the equation zn = 1 or zn − 1 = 0 The roots are given by

µ ¶ 2kπ 2kπ zk = cos + i sin , n n µ ¶ 2π 2π k = cos + i sin , n n

k = 0, 1, 2, . . . , n − 1 k = 0, 1, 2, . . . , n − 1

2π 2π Setting cos + i sin = α, the nth roots of unity are 1, α, α2 , . . . , αn−1 . n n The sum of the roots, 1 + α + α2 + . . . + αn−1 =

1 − αn =0 1−α

(since α is a root of z n − 1 = 0)

while the product of the roots is 1 · α · α2 · · · · · αn−1 = α1+2+3+...+(n−1) "µ ¶ n #n−1 ³ ´(n−1) 2π 2 2π (n−1)n/2 n/2 + i sin =α = α = cos n n = (cos π + i sin π)n−1

applying De Moivre’s formula

n−1

= (−1) Anant Kumar


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Geometry of complex numbers

Thus, z n − 1 = (z − 1)(z − α)(z − α2 ) · · · (z − αn−1 ).

3 3.1

Geometry of complex numbers Euler’s formula. Rotation as complex multiplication

Euler’s formula expresses an exponential function with an imaginary exponent in terms of trigonometric functions: ei ϕ = cos ϕ + i sin ϕ (18) Replacing ϕ by −ϕ in (18), we obtain e−i ϕ = cos ϕ − i sin ϕ

(19)

From (18) and (19), we find cos ϕ and sin ϕ:  ei ϕ + e−i ϕ   cos ϕ = 2 ei ϕ − e−i ϕ   sin ϕ = 2i

(20)

These formulae are used in particular to express powers of cos ϕ and sin ϕ and their products in terms of the sine and cosine of multiple arcs. Example 2. µ 2

cos y =

ei y + e−i y 2

¶2 =

¢ 1 ¡ i 2y e + 2 + e−i 2y 4

1 [(cos 2y + i sin 2y) + 2 + (cos 2y − i sin 2y)] 4 1 1 = (2 cos 2y + 2) = (1 + cos 2y) 4 2 =

Example 3. µ 2

2

cos ϕ sin ϕ =

ei ϕ + e−i ϕ 2

¶2 µ

ei ϕ − e−i ϕ 2i

¶2

¡ i 2ϕ ¢2 e − e−i 2ϕ 1 1 = = − cos 4ϕ + 4 · 4i 2 8 8

The exponential form of a complex number. Let us represent a complex number in the trigonometric form z = r(cos ϕ + i sin ϕ) where r is the modulus of the complex number and ϕ is the argument. By Euler’s formula, cos ϕ + i sin ϕ = ei ϕ . Thus, the complex number can be represented in the so–called exponential form: z = rei ϕ (21) By the properties of exponentials, it is easy to operate on complex number in exponential form. For example, for two given z1 = r1 ei ϕ1 and z2 = r2 ei ϕ2 , we have z1 z2 = r1 ei ϕ1 · r2 ei ϕ2 = r1 r2 ei (ϕ1 +ϕ2 ) ,

and

z1 r1 ei ϕ1 r1 i (ϕ1 −ϕ2 ) = = e i ϕ z2 r2 e 2 r2

Interpretation of ei ϕ as rotation. Consider a real number a. It can be treated as a complex −→ number and hence represented as a vector OA as shown in Figure. Now keeping the tail fixed at Anant Kumar


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the origin, let us rotate it in the anticlockwise direction by an angle ϕ. The tip of the vector moves in a circular path of radius a and ultimately comes to rest at the position P of the complex plane. Since P lies on the complex plane, it must represented a complex number z whose modulus is the length of OP = a and whose argument is the angle ϕ, i.e. it represents the complex number z = aei ϕ . But then z can be interpreted as the product of the real number a and the unit– modular complex number number ei ϕ . Thus, multiplication of a real number by the number ei ϕ amounts to the rotation of the vector representing the real number in the complex plane in the anticlockwise direction by an angle ϕ. Further, taking the multiplication of two complex number z

y

y P (a ·

z

zz1

ei ϕ ) r|z

a

|

1

ϕ

z1

ϕ O

a

(a 0)

x

x

O

(a)

(b)

Figure 5: (a) Multiplying a real number with ei ϕ has the effect of rotating the vector representing the real

number in the anticlockwise sense by an angle ϕ. (b) z1 is multiplied by z = rei ϕ resulting in rotation as well as scaling.

into account, we can generalize to the following statement: Multiplying a complex number z1 by another complex number z = rei ϕ has the effect of scaling the length of the vector representing z1 by a factor r and simultaneously rotating it in the anticlockwise direction by an angle ϕ. Since, i = ei π/2 , therefore, multiplying any number by i has the effect of rotating its vector representation in the anticlockwise sense by 90◦ , while multiplying it by −1 = i2 = i · i rotates in the anticlockwise sense by 180◦ .

3.2

Conjugate

For the complex number z, we know that z differs z by only the sign of the argument. That means that the conjugate of rei ϕ is just re−i ϕ . But that also means that z is just the reflection of z on the real line. It is easy to verify that z + z = 2Re (z)

and

z − z = 2i Im (z)

(22)

z+w =z+w

and

zw = zw

(23)

From the above relations, we see that if z = z, then Im (z) = 0, i.e. z is purely real (only a real number equals its conjugate). Similarly, if z + z = 0, then z must be purely imaginary.

3.3

Modulus — the distance function

We have already noted that for a complex number z, the modulus is given by the relation |z|2 = zz, and |z1 − z2 | represents the distance between the points in the complex plane representing the numbers z1 and z2 . Looked in this manner, |z| = |z − 0| represents the distance of the point z Anant Kumar


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from the origin. The following are some basic properties of the modulus, which can be easily verified |z| = |z| = | − z| = | − z| |zw| = |z||w| ¯z¯ |z| ¯ ¯ ¯ ¯= w |w|

(24) (25)

w, z ∈ C

(26)

A fundamental property of any distance function is that it satisfies the triangle’s inequality: for any two complex numbers z and w, we have |z + w| ≤ |z| + |w|

(27)

This is called the triangle inequality because, if we represent z and w by, respectively, the points P and Q in the complex plane, (27) says that the sum of length of one side of the triangle OP Q (where O is the origin) is less than the sum of the lengths of the other two sides. Or, the shortest distance between two points is a straight line. In order to prove (27), first observe that, for any z ∈ C, −|z| ≤ Re (z) ≤ |z|,

and

− |z| ≤ Im (z) ≤ |z|

(28)

Hence, Re (zw) ≤ |zw| = |z||w|. Thus, |z + w|2 = (z + w)(z + w) = zz + zw + wz + ww = |z|2 + zw + zw + |w|2 = |z|2 + 2Re (zw) + |w|2 ≤ |z|2 + 2|z||w| + |w|2 = (|z| + |w|)2 ⇒

|z + w| ≤ |z| + |w|

On encountering an inequality one should ask for necessary and sufficient conditions that the equality holds. From looking at a triangle and considering the geometrical significance of (27), we are led to consider the condition for equality that the two complex numbers must be collinear with the origin, that for some t ∈ R, t ≥ 0, we must have z = tw (or w = tz, if w = 0). In fact, if we look at the proof of this inequality, we see that a necessary and sufficient condition for |z + w| = |z| + |w| is that |zw| = Re (zw). This means, that zw must be a non–negative real number : zw ≥ 0 z Multiplying this by w/w, we get |w|2 ≥ 0 if w 6= 0. If w ¶ µ z 1 |w|2 t= |w|2 w then t ≥ 0 and z = tw. By induction, we also get |z1 + z2 + . . . + zn | ≤ |z1 | + |z2 | + . . . + |zn | Also useful is the inequality,

Anant Kumar

¯ ¯ ¯ ¯ |z| − |w| ¯ ¯ ≤ |z − w|


(29)

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Mob. No. 9932347531

13


µ 3.4

Interpretation of arg

z1 − z0 z2 − z0

¶ µ

¶ z1 By the rule on quotient of two complex numbers z1 and z2 , we know that arg gives the z2 angle between the lines joining the origin O and the representation of z1 and z2 . This angle is measured as one “walks from the denominator to the numerator.” Actually, we can interpret the ratio z1 /z2 as (z1 − 0)/(z2 − 0). In this form, it is clear that z1 − 0 is the directed line from the origin to µ the point ¶ z1 and similarly for z2 . Thus, we can z1 − z0 generalize the above interpretation that arg represents the angle between the join of z2 − z0 z0 to z1 and the join of z0 to z2 , as one walks from the denominator to the numerator. To prove this, let z0 , z1 , and z2 be represented, respectively, as points A, B, and C in the complex plane (Figure 6(a)). Further, let P and Q represent the complex numbers z1 − z0 and z2 − z0 , z1 − z0 respectively. It is obvious that 4ABC is congruent to 4OP Q. In triangle, we interpret z2 − z0 Im

z

Im

C(z2 )

z z1

B(z1 )

θ

θ

A(z0 )

z4

Q(z2 − z0 )

³ z3

P (z1 − z0 )

θ

Re

O

θ = arg z2

´

Re

O

(a)

z1 −z2 z2 −z4

(b)

Figure 6: (a) Interpretation of the argument of (z1 − z0 )/(z2 − z0 ). (b) that of (z1 − z2 )/(z3 − z4 ) as the complex number whose modulus is

OP while the argument is θ. Thus OQ

OP AB |z1 − z0 | z1 − z0 = (cos θ + i sin θ) = (cos θ + i sin θ) = (cos θ + i sin θ) z2 − z0 OQ AC |z2 − z0 | z1 − z2 is the angle between the join z3 − z4 of z2 to z1 and z4 to z3 as one walks from the denominator to the numerator (see Figure 6(b)). By a similar token, the argument of the complex number

♣ Remark 6. Three points z1 , z2 , and z3 are collinear if and only if, µ ¶ z1 − z3 z1 − z3 arg = 0 or π ⇔ is purely real. z2 − z3 z2 − z3 ♣ Remark 7. The line joining z1 and z2 is perpendicular to the line joining z3 and z4 if and only if µ ¶ z1 − z2 π z1 − z3 arg =± ⇔ is purely imaginary. z3 − z4 2 z2 − z3

3.5

Lines in the complex plane

Let L denote a line in C. We know that a straight line in a plane is uniquely determined two specified pieces of information. Let us say that we have been specified two complex numbers z1 and z2 on L. Now, if z be an arbitrary point on L, then µ ¶ z − z2 z − z2 arg = 0 or π ⇒ = purely real = t (say), where − ∞ ≤ t ≤ ∞ z1 − z2 z1 − z2 Anant Kumar


Mob. No. 9932347531

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Solving the last equation for z, we obtain the equation for a straight line in the parametric form: Parametric form of straight line: Here t is called the parameter. Again, since z − z2 z − z2 = z1 − z2 z1 − z2

⇒

z = tz1 + (1 − t)z2 ,

−∞ ≤ t ≤ ∞

(31)

z − z2 is purely real, we have z1 − z2

z(z 1 − z 2 ) − z(z1 − z2 ) + z1 z 2 − z 1 z2 = 0

Thus we obtain the equation in two point form: Two point form of a straight line: z(z 1 − z 2 ) − z(z1 − z2 ) + z1 z 2 − z 1 z2 = 0

(32)

Finally, transform left side the above equality as z(z 1 − z 2 ) − z(z1 − z2 ) + z1 z 2 − z 1 z2 = z(z 1 − z 2 ) − z(z1 − z2 ) + z1 z 2 − z1 z 2 = z(z 1 − z 2 ) − z(z1 − z2 ) + 2i Im (z1 z 2 ) 1 = z(z 1 − z 2 ) + zi (z1 − z2 ) + 2 Im (z1 z 2 ) (dividing by i ) i£ ¤ £ ¤ = z − i (z 1 − z 2 ) + z i (z1 − z2 ) + 2 Im (z1 z 2 ) Setting a = i (z1 − z2 ) and b = 2 Im (z1 z 2 ), we obtain the general equation: General form of straight line:

az + az + b = 0,

a ∈ C, b ∈ R

(33)

The complex slope of the line joining z1 and z2 in the complex plane is defined as µ=

z1 − z2 z1 − z2

(34)

Thus, for the line represented by (33), the complex slope becomes µ=−

a a

(35)

Further, two lines in the complex plane having complex slopes µ1 and µ2 are (i) parallel, if µ1 = µ2 , (ii) perpendicular, if µ1 + µ2 = 0.

3.6

Circles in the complex plane

Given the center of the circle, z0 , and its radius, r, we note that if z be any arbitrary point on the circle, its distance from the center must be the same as the radius: Circle:

|z − z0 | = r

(36)

Transform as follows: r2 = |z − z0 |2 = (z − z0 )(z − z 0 ) = zz − z0 z − z 0 z + z0 z 0 Setting a = −z0 , and b = z0 z 0 − r2 , we obtain the general equation of the circle as zz + az + az + b = 0, Anant Kumar

a ∈ C, b ∈ R


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15


√ having center at the point represented by −a and radius as aa − b. Finally, if the end points of a diameter be given as A(z1 ) and B(z2 ), then the equation of the circle is (z − z1 )(z − z 2 ) + (z − z 1 )(z − z2 ) = 0

(38)

The above equation follows if we take into account the fact that if P (z) be an arbitrary point on the required circle, then µ ¶ π z − z1 =± , arg z − z2 2 since ∠AP B = 90◦ , being an angle in a semi–circle. Therefore, the complex number purely imaginary, and so its real part is zero:

z − z1 is z − z2

z − z1 z − z 1 + =0 z − z2 z − z 2 Rearranging the above expression, (38) follows. Example 4 (Section formula). The points A and B on the complex plane have affixes z1 and z2 . Find the affix of the point P that divides the join of A and B in the ratio m : n, i.e. AP/P B = m/n. Solution: Let z be the required affix. Then the argument of (z − z1 )/(z2 − z) is zero. Thus z − z1 |z − z1 | |z − z1 | PA m = (cos 0 + i sin 0) = = = z2 − z |z2 − z| |z2 − z| BP n Solving for z, we obtain z=

Anant Kumar

mz2 + nz1 m+n


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Mob. No. 9932347531

Complex numbers

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