Mathematics COMPLEX NUMBERS - Introduction , Algebra , Modulus and Conjugate of Complex Numbers
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### Topics Covered

star Introduction
star Complex Numbers
star Addition of two complex numbers
star Difference of two complex numbers
star Multiplication of two complex numbers
star Division of two complex numbers
star Power of i
star The square roots of a negative real number
star Identities
star The Modulus and the Conjugate of a Complex Number

### Introduction

● We have seen that the equation color(blue)(x^2 + 1 = 0) has no real solution as x^2 + 1 = 0 gives x^2 = – 1 and square of every real number is non-negative.

● So, we need to extend the real number system to a larger system so that we can find the solution of the equation x^2 = – 1.

● In fact, the main objective is to solve the equation color(blue)(ax^2 + bx + c = 0), where color(blue)(D = b^2 – 4ac < 0), which is not possible in the system of real numbers.

### Complex Numbers

● Let us denote color(red)(sqrt (−1)) by the symbol color(red)(i). Then, we have color(blue)(i^2 = −1) . This means that i is a solution of the equation x^2 + 1 = 0.

● A number of the form a + ib, where a and b are real numbers, is defined to be a complex number.
color(red)("For example"), 2 + i3, (– 1) + i sqrt(3) are color(blue)("complex numbers.")

● For the complex number color(blue)(z = a + ib),

color(blue)(a) is called the color(blue)("real part"), denoted by color(blue)("Re z) and color(blue)(b) is called the color(blue)("imaginary part") denoted by color(blue)(Im z) of the complex number z.

For example, if z = 2 + i5, then Re z = 2 and Im z = 5.

● Two complex numbers z_1 = a + ib and z_2 = c + id are equal if a = c and b = d.
Q 3028801701

If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

Solution:

We have
4x + i (3x – y) = 3 + i (–6) ... (1)
Equating the real and the imaginary parts of (1), we get

4x = 3, 3x – y = – 6,

which, on solving simultaneously, give x = 4/3  and y = 33/4

### Addition of two complex numbers

Let z_1 = a + ib and z_2 = c + id be any two complex numbers.

● Then, the sum z_1 + z_2 is defined as follows: z_1 + z_2 = (a + c) + i (b + d), which is again a complex number.
For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8

● The addition of complex numbers satisfy the following properties:

color(blue)("(i) The closure law") The sum of two complex numbers is a complex number,
i.e., z_1 + z_2 is a complex number for all complex numbers z_1 and z_2.

color(blue)("(ii) The commutative law") For any two complex numbers z_1 and z_2,

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)(z_1 + z_2 = z_2 + z_1)

color(blue)( "(iii) The associative law") For any three complex numbers z_1, z_2, z_3,

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red)((z_1 + z_2) + z_3 = z_1 + (z_2 + z_3)).

color(blue)("(iv) The existence of additive identity ") There exists the complex number 0 + i 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number z, color(red)(z + 0 = z.)

color(blue)"(v) The existence of additive inverse" To every complex number z = a + ib, we have the complex number – a + i(– b) (denoted as – z), called the additive inverse or negative of z. We observe that z + (–z) = 0 (the additive identity).

### Difference of two complex numbers

● Given any two complex numbers z_1 and z_2, the difference z_1 – z_2 is defined as follows:

color(red)(z_1 – z_2 = z_1 + (– z_2).)

### Multiplication of two complex numbers

Let z_1 = a + ib and z_2 = c + id be any two complex numbers.

● Then, the product z_1 z_2 is defined as follows: z_1 z_2 = (ac – bd) + i(ad + bc)
For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28

● The multiplication of complex numbers possesses the color(red)("following properties"),

color(blue)"(i) The closure law" The product of two complex numbers is a complex number, the product z_1 z_2 is a complex number for all complex numbers z_1 and z_2.

color(blue)"(ii) The commutative law" For any two complex numbers z_1 and z_2,

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){z_1 z_2 = z_2 z_1}
.
color(blue)"(iii) The associative law" For any three complex numbers z_1, z_2, z_3,

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(red){ (z_1 z_2) z_3 = z_1 (z_2 z_3)}.

color(blue)"(iv) The existence of multiplicative identity" There exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that color(red)(z.1 = z), for every complex number z.

color(blue)"(v) The existence of multiplicative inverse" For every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0),

we have the complex number a/(a^2+b^2)+i(-b)/(a^2+b^2) (denoted by 1/z or z^(-1) called the multiplicative inverse of z
such that z. 1/z = 1  (the multiplicative identity).

color(blue) "(vi) The distributive law" For any three complex numbers z_1, z_2, z_3,
(a) \ \ color(red)(z_1 (z_2 + z_3) = z_1 z_2 + z_1 z_3)

(b) \ \ color(red)((z_1 + z_2) z_3 = z_1 z_3 + z_2 z_3)

### Division of two complex numbers

Given any two complex numbers z_1 and z_2,

where z_2 ≠ 0 , the quotient z_1/z_2 is defined by  color(red)(z_1/z_2=z_1* 1/z_2)

### Power of i

we know that

● color(blue)(i^3) = i^2i = (−1) i = −i ,

●  color(blue)(i^4) = (i^2)^2 = (−1)^2 =1

●  color(blue)(i^5) = (i^2)^2* i = (−1 )^2 i = i ,

● color(blue)(i^6) = (i^2)^3 = (−1)^3 =-1 , etc

In general, for any integer k, i^(4k) = 1, i^(4k + 1) = i,  i^(4k + 2) = -1, i^(4k +3) = -i

### The square roots of a negative real number

● Note that color(blue)(i^2 = –1) and color(blue)(( – i)^2) = i^2 = – 1

Therefore, the square roots of – 1 are i, – i.

● However, by the symbol  sqrt(−1) , we would mean i only.

Now, we can see that color(blue)(i) and color(blue)(–i) both are the solutions of the equation color(blue)(x^2 + 1 = 0) or x^2 = –1.

● Similarly

color(blue)((sqrt3 i)^2) = (sqrt3)^2 i^2 = 3(-1) = -3

color(blue)((-sqrt3 i)^2) = (-sqrt3)^2 i^2 = -3

The square roots of color(red)(–3) are color(red)(sqrt (3) * i) and color(red)(− sqrt(3) * i)

● Generally, if a is a positive real number, sqrt(−a) = sqrt(a)* sqrt(−1 )= sqrt(a)* i ,.

We already know that color(blue)(sqrt(a) ×sqrt( b) = sqrt(ab)) for all positive real number a and b.

This result also holds true when either a > 0, b < 0 or a < 0, b > 0.

● What if color(red)(a < 0, b < 0 \ \?)  Let us examine.

Note that

i^2 = sqrt(−1) * sqrt(−1) = sqrt((−1) (−1))  (by assuming a × b = ab for all real numbers) = sqrt(1) = 1, which is a contradiction to the fact that i ^2 = − 1.

● Therefore, color(red)(sqrt(a) × sqrt(b) != sqrt(ab)) if both a and b are negative real numbers.

Further, if any of a and b is zero, then, clearly, sqrt(a) × sqrt(b) = sqrt(ab) = 0.

### Identities

● We prove the following identity

color(blue)((z_1 + z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2 ,) for all complex numbers z_1 and z_2.

color(red)"Proof" We have, (z_1 + z_2)^2 = (z_1 + z_2) * (z_1 + z_2),
= (z_1 + z_2) z_1 + (z_1 + z_2) z_2 (Distributive law)
= z_1^2 + z_2z_1 + z_1z_2 + z_2^2 (Distributive law)
= z_1^2 + z_2z_1 + z_1z_2 + z_2^2 (Commutative law of multiplication )
=z_1^2 + z_2^2 + 2z_1z_2 ,

● Similarly, we can prove the following identities:
(i) color(blue)((z_1 - z_2)^2 = z_1^2 + z_2^2 - 2z_1z_2)

(ii) color(blue)(( z_1 + z_2)^3 = z_1^3 + 3z_1^2 z_2 + 3z_1z_2^2 + z_2^3)

(ii) color(blue)(( z_1 - z_2)^3 = z_1^3 - 3z_1^2 z_2 + 3z_1z_2^2 - z_2^3)

(iv) color(blue)(z_1^2 – z_2^2 = (z_1 + z_2) *( z_1 – z_2))
Q 3048801703

Express the following in the form of a + bi:

(i) (-5i)(1/8 i)

(ii) (-i) (2i) (-1/8 i)^3

Solution:

(i) (-5i)(1/8 i) = (-5)/8 i^2 = (-5)/8 i^2 = (-5)/8 (-1) = 5/8+5/8 i 0

(ii) (-i) (2i) (-1/8 i)^3 = 2 xx 1/(8xx8xx8) xx i^5 = 1/(256) (i^2)^2 i = 1/256 i
Q 3058801704

Express (5 – 3i)^3 in the form a + ib.

Solution:

We have, (5 – 3i)^3 = 53 – 3 × 5^2 × (3i) + 3 × 5 (3i)^2 – (3i)^3
= 125 – 225i – 135 + 27i = – 10 – 198i.
Q 3018001800

Express (- sqrt3+sqrt(-2) ) (2 sqrt3-i) in the form of a + ib

Solution:

We have, (- sqrt3 + sqrt(-2) ) (2 sqrt3 -i) = ( - sqrt3 + sqrt2 i)( 2 sqrt3-i)

= -6+sqrt3i+2 sqrt6i - sqrt2 i^2 = (-6+sqrt2) + sqrt3(1+2 sqrt2) i

### The Modulus and the Conjugate of a Complex Number

● Let z = a + ib be a complex number.

Then, color(blue)("the modulus of z, denoted by") color(red)(| z |), is defined to be the non-negative real number sqrt(a^2+B^2) ,

i.e. color(blue)(| z | = sqrt(a^2+b^2))

● The conjugate of z, denoted as color(blue)(barz) , is the complex number a – ib, i.e., barz = a – ib.

For example | 3+i | = sqrt(3^2+1^2) = sqrt(10) , | 2-5i| = sqrt(2^2+(-5)^2) = sqrt(29) and bar(3+i) = 3-i , bar(2-5i) = 2+5i , bar(-3i-5) = 3i-5

● Observe that the multiplicative inverse of the non-zero complex number z is given by

color(blue)(z^(-1) )= 1/(a+ib) = a/(a^2+b^2) +i (-b)/(a^2+b^2) = (a-ib)/(a^2+b^2) =color(blue)( barz/(|z|^2)) or color(red)(z barz = | z|^2)

● For any two complex numbers z_1 and z_2 , we have

(i) color(blue)( | z_1 z_2 | = | z_1 | | z_2 |)

(ii) color(blue)(| z_1/z_2 | = (| z_1 |)/(| z_2 |)) provided  | z_2 | ne 0

(iii) color(blue)(bar(z_1 z_2) = barz_1 barz_2)

(iv) color(blue)(bar(z_1 pm z_2) = barz_1 pm barz_2)

(v) color(blue)(bar ((z_1/z_2)) = barz_1/barz_2) provided z_2 ne 0
Q 3018112000

Find the multiplicative inverse of 2 – 3i.

Solution:

Let z = 2 – 3i

Then z = 2 + 3i and z^2 = 2^2 + ( − 3)^2 =13
Therefore, the multiplicative inverse of 2 − 3i is given by

z^(-1) = barz/(| z|^2) = (2+3i)/(13) = 2/13+3/13 i

The above working can be reproduced in the following manner also,

z^(-1) = 1/(2-3i) = (2+3i)/{(2-3i) (2+3i)}

= (2+3i)/(2^2-(3i)^2) = (2+3i)/(13) = 2/13+3/13 i
Q 3008112008

Express the following in the form a + ib

(i) (5+sqrt2i)/(1- sqrt2 i)

(ii) i^(-35)

Solution:

(i) We have, (5+ sqrt2 i)/(1- sqrt2 i) = ( 5+ sqrt2 i)/(1-sqrt2i) xx ( 1+ sqrt2 i)/(1+ sqrt2 i)

= (5+5 sqrt2 i + sqrt2i -2)/(1- ( sqrt2i)^2)

= (3+6 sqrt2 i)/(1+2) = (3 (1+2sqrt2i))/3 = 1+2 sqrt2i

(ii) i^(-35) = 1/i^(35) = 1/{(i^2)^(17)i} = 1/(-i) xx i/i = i/(-i^2) = i`