Diploma_Semester-II_Advanced Mathematics_Complex number

Complex number

Course- Diploma Semester-II

Subject- Advanced Mathematics Unit- II

RAI UNIVERSITY, AHMEDABAD

Unit-II COMPLEX NUMBER


Introduction— Indian mathematician Mahavira (850 A.D.) was first to mention in his work 'Ganitasara Sangraha'; 'As in nature of things a negative (quantity) is not a square (quantity), it has, therefore, no square root'. Hence there is no real number 푥 which satisfies the polynomial equation 푥 + 1 = 0. A symbol√−1, denoted by letter “i” was introduced by Swiss Mathematician, Leonhard Euler (1707-1783) in 1748 to provide solutions of equation 푥 + 1 = 0. “i” was regarded as a fictitious or imaginary number which could be manipulated algebraically like an ordinary real number, except that its square was – 1. The letter “i” was used to denote√−1 , possibly because “i” is the first letter of the Latin word 'imaginarius'. 2.1Definition— A number of the form 푧 = 푥 + 푖푦, [푥,푦 ∈ 푅 & 푖 = √−1 ], is called a complex number.

Figure 2.01 represents the complex plane. It consists of a Real axis and an Imaginary axis. 2. 2 Real & Imaginary part of a complex number— Let 푧 = 푥 + 푖푦 is a complex number. Then it’s real part & imaginary part is given by— 푅푒(푍) = 푥 푎푛푑 퐼푚(푍) = 푦 2. 3 Representation of a complex number 풛 = 풙 + 풊풚 – A complex number 푧 = 푥 + 푖푦 can be represented on the co-ordinate as given below—

Fig - 2.01

This system of representing complex number is also called as Argand diagram.



2. 4 Modulus of a complex number— Let 풛 = 풙 + 풊풚 is a complex number, then its modulus is given by— |푧|= 푥 + 푦 . Example— Find the modulus of the complex number 퐳 = ퟑ + ퟒ퐢. Solution— Given complex number is z = 3 + 4i. Since, modulus of z = x + iy is given by |푧|= 푥 + 푦 Hence, required modulus of z is |푧|=√3 + 4 =√9 + 16 = √25 ∴ |푧| = 5 2. 5 Argument of a complex number— Let 풛 = 풙 + 풊풚 is a complex number, then its argument is given by—

arg(푧) = 휃 = tan . Note— 휃 is measured in radians and positive in the counterclockwise sense. For 풛 = ퟎ , 퐚퐫퐠(풛) is not defined. The value of 휃 that lies in the interval −휋 ≤ 휃 ≤ 휋 is called as Principal value of the argument of 푧(≠ 0).

Hence, −흅 ≤ 퐚퐫퐠(풛) ≤ 흅 Example— Find the argument of the complex number 퐳 = ퟏ + 퐢. Solution— given complex number is z = 1 + i. Since, argument of z = x + iy is given by arg(푧) = tan and the point (1,1) lies in first quadrant. Hence, arg(푧) = tan = tan (1) = 흅

ퟒ

Example— Find the argument of the complex number 퐳 = −ퟏ + 퐢. Solution— given complex number is z = −1 + i. Since, argument of z = x + iy is given by arg(푧) = tan and the point (−1,1) lies in second quadrant. Hence, arg(푧) = tan = tan (−1) = 흅 − 흅

ퟒ= ퟑ흅

ퟒ

Example— Find the argument of the complex number 퐳 = −ퟏ − 퐢. Solution— given complex number is z = −1 − i. Since, argument of z = x + iy is given by arg(푧) = tan and the point (−1,−1) lies in third quadrant. Hence, arg(푧) = tan = tan (1) = −흅 + 흅

ퟒ= ퟑ흅

ퟒ

Example— Find the argument of the complex number 퐳 = ퟏ − 퐢. Solution— given complex number is z = 1 − i.

Fig - 2.02



Since, argument of z = x + iy is given by arg(푧) = tan and the point (1,−1) lies in fourth quadrant. Hence, arg(푧) = tan = tan (−1) = − 흅

ퟒ

2.6 Cartesian representation of a complex number— A complex number of the form 푧 = 푥 + 푖푦 is known as Cartesian representation of complex number. This can also be written in paired form as (푥,푦). Cartesian representation of complex number 푧 = 푥 + 푖푦 is shown in Fig- 1.02. 2.7 Polar representation of a complex number— A complex number of the form 푧 = 푥 + 푖푦 can be represented in polar form as 푧 = 푟(푐표푠휃 +푖푠푖푛휃). Where 푟 and 휃 is given by—

푟 = 푥 + 푦 and

휃 = tan푦푥

This is shown in Fig- 2.03. 2.8 Euler representation of a complex number— A complex number of the form 푧 = 푥 + 푖푦 or 푧 = 푟(푐표푠휃 + 푖푠푖푛휃) can be represented in “Euler form” form as---

푧 = 푟푒 Where 푟 and 휃 is given by— 푟 = 푥 + 푦 , 휃 = tan This is shown in Fig- 2.03. 2.9 Conversion of a complex number given in cartesian system to polar system— If 푧 = 푥 + 푖푦 is a complex number given in Cartesian system then it can be written in polar form by writing— 푧 = 푟(푐표푠휃 + 푖푠푖푛휃) Where 푟 = 푥 + 푦 and 휃 = tan . Example— Convert the following complex numbers in polar form— (풂) ퟑ + ퟒ풊 (풃) ퟏ + 풊 (풄) − ퟏ + 풊 (풅) ퟏ + √ퟑ풊 Solution— (푎) Given complex number is 푧 = 3 + 4푖, Hence, 푟 = √3 + 4 = √9 + 16 = √25 = 5 ∴ 푟 = 5 and 휃 = tan = tan

Fig - 2.03



Thus, 푧 = 3 + 4푖 can be written in polar form as 푧 = 5(푐표푠휃 + 푖푠푖푛휃), where 휃 = tan (푏) Given complex number is 푧 = 1 + 푖, Hence, 푟 = √1 + 1 = √1 + 1 = √2 ∴ 푟 = √2 and 휃 = tan = tan =

Thus, 푧 = 1 + 푖 can be written in polar form as 푧 = √2(푐표푠 + 푖푠푖푛 ) (푐) Given complex number is 푧 = −1 + 푖, Hence, 푟 = (−1) + 1 = √1 + 1 = √2 ∴ 푟 = √2 and 휃 = tan = tan =

Thus, 푧 = −1 + 푖 can be written in polar form as 푧 = √2(푐표푠 + 푖푠푖푛 ) (푑) Given complex number is 푧 = 1 + √3푖,

Hence, 푟 = (1) + (√3) = √1 + 3 = √4 = 2 ∴ 푟 = 2 and

휃 = tan = tan √ =

Thus, 푧 = 1 + √3푖 can be written in polar form as 푧 = 2(푐표푠 + 푖푠푖푛 ) 2.10 Conversion of a complex number given in polar system to cartesian system —

Example— Convert the followings complex numbers in Cartesian form—

(푎) 푒 (푏) √2 푒 (푐)√3푒 (푑) √2푒 Solution— (푎) Given complex number is 푒 . Here, 푟 = 1 푎푛푑 휃 = , let 푧 = 푥 + 푖푦 is Cartesian representation.

Then, 푥 = 푟푐표푠휃 = 1. 푐표푠 = 0 and

푦 = 푟푠푖푛휃 = 1. 푠푖푛휋2

= 1

Fig - 2.04

If 푧 = 푟(cos휃 + 푖푠푖푛 휃) is a complex number given in Polar system then it can be written in Cartesian system as— 푧 = 푥 + 푖푦 Where 푥 = 푟푐표푠휃 and 푦 = 푟푠푖푛휃.



Thus, 푧(1, ) can be written in Cartesian form as 푧 = 0 + 푖

(푏) Given complex number is √2 푒 Here, 푟 = √2 푎푛푑 휃 = , let 푧 = 푥 + 푖푦 is Cartesian representation.

Then, 푥 = 푟푐표푠휃 = √2. 푐표푠 = 1 and 푦 = 푟푠푖푛휃 = √2. 푠푖푛 = 1

Thus, √2 푒 can be written in Cartesian form as 푧 = 1 + 푖 (푐) Given complex number is √3푒 Here, 푟 = √3 푎푛푑 휃 = , let 푧 = 푥 + 푖푦 is Cartesian representation.

Then, 푥 = 푟푐표푠휃 = √3. 푐표푠 = √ and

푦 = 푟푠푖푛휃 = √3. 푠푖푛 =

Thus, √3푒 can be written in Cartesian form as 푧 = √ + 푖

(푑) Given complex number is √2푒 Here, 푟 = √2 푎푛푑 휃 = , let 푧 = 푥 + 푖푦 is Cartesian representation.

Then, 푥 = 푟푐표푠휃 = √2. 푐표푠 = −1 and

푦 = 푟푠푖푛휃 = √2. 푠푖푛3휋4

= 1

Thus, √2푒 can be written in Cartesian form as 푧 = −1 + 푖 2.10 Conjugate of a complex number—

Example— Find the conjugate of the complex number 퐳 = ퟐ + ퟑ퐢. Solution— Given complex number is z = 2 + 3i. Since, complex conjugate of z = x + iy is given by 풛 = 풙 − 풊풚. Hence, required complex conjugate is 풛 = ퟐ − ퟑ풊. 2.11 Addition of two complex numbers—

Fig - 1.05

Let a complex number is given by 풛 = 풙 + 풊풚, then the conjugate of complex number 푧 is denoted by 푧̅ and it is given by 풛 = 풙 − 풊풚. A complex number 풛 = 풙 + 풊풚 and its conjugate 풛 = 풙 − 풊풚 is represented in Argand plane as shown in Fig- 2.05. Hence,

⇒푹풆(풛) = 풙 = ퟏ

ퟐ(풛 + 풛) & 푰풎(풛) = 풚 = ퟏ

ퟐ풊(풛 − 풛)



Let 푧 = 푥 + 푖푦 and 푧 = 푥 + 푖푦 are two complex numbers, then addition of these two complex numbers 푧 and 푧 is given by—

푍 = (푥 +푥 ) + 푖(푦 + 푦 ) Example— Find the addition of the complex numbers 풛ퟏ = ퟐ + ퟑ풊 and 풛ퟐ = ퟒ + ퟐ풊 . Solution— given complex numbers are 푧 = 2 + 3푖 and 푧 = 4 + 2푖 . Hence, 푧 + 푧 =(2 + 4) + 푖(3 + 2) ∴ 푍 = 6 + 5푖 2.12 Subtraction of two complex numbers— Let 푧 = 푥 + 푖푦 and 푧 = 푥 + 푖푦 are two complex numbers, then subtraction of complex numbers 푧 from 푧 is given by—

푍 = (푥 −푥 ) + 푖(푦 − 푦 ) Example— Subtract the complex numbers 풛ퟏ = ퟐ + ퟑ풊 from 풛ퟐ = ퟒ + ퟐ풊 . Solution— given complex numbers are 푧 = 2 + 3푖 and 푧 = 4 + 2푖 . Hence, 푧 − 푧 =(4 − 2) + 푖(2 − 3) ∴ 푍 = 2 − 푖 2.13 Multiplication of two complex numbers— Let 푧 = 푥 + 푖푦 and 푧 = 푥 + 푖푦 are two complex numbers, then multiplication of complex numbers 푧 and 푧 is given by—

푍 = (푥 푥 −푦 푦 ) + 푖(푥 푦 + 푥 푦 ) Example—Multiply the complex numbers 풛ퟏ = ퟑ + ퟒ풊 and 풛ퟐ = ퟒ + ퟑ풊 . Solution— Given complex numbers are 푧 = 3 + 4푖 and 푧 = 4 + 3푖 . Hence, (푧 푧 ) =(3 + 4푖)(4 + 3푖) = (12 − 12) + 푖(9 + 16) ∴ 푍 = 25푖 2.14 Division of two complex numbers— Let 푧 = 푥 + 푖푦 and 푧 = 푥 + 푖푦 are two complex numbers, then division of complex numbers 푧 by 푧 is given by—

푍 =푧푧

=푥 + 푖푦푥 + 푖푦

Now, multiplying and devide by 푧 = 푥 − 푖푦 . Hence,

푍 =푧푧

= (푥 + 푖푦푥 + 푖푦

)(푥 − 푖푦푥 − 푖푦

)

Now, multiplying numerator and denominator,

푍 =푧푧

=(푥 푥 +푦 푦 ) + 푖(푥 푦 − 푥 푦 )

(푥 + 푦 )

∴ 푍 =푧푧

=(푥 푥 +푦 푦 )

(푥 + 푦 )+ 푖

(푥 푦 − 푥 푦 )(푥 + 푦 )



Example—Divide the complex numbers 풛ퟏ = ퟏ + 풊 by 풛ퟐ = ퟑ + ퟒ풊 . Solution— given complex numbers are 푧 = 1 + 푖 and 푧 = 3 + 4푖 . Hence,

푍 =푧푧 =

3 + 4푖1 + 푖

Now, multiplying and devide by 푧 = 1 − 푖 Hence, 푍 = =

= ( ) ( ) =

∴ 푍 = = + 푖

2.15 De Moivre’s Theorem— In mathematics, De Moivre's formula or De Moivre's identity states that for any complex number 푧 = 푒 and integer "푛" it holds that—

(푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃) Example— Solve by using De Moivre’s Theorem—

(a) (풄풐풔휽 + 풊풔풊풏휽)ퟐ (풃) (ퟏ + 풊)ퟓ (풄) ퟏퟐ

+ √ퟑퟐ풊ퟏퟎ

(풅) 풄풐풔휽 풊풔풊풏휽풄풐풔휽 풊풔풊풏휽

ퟏퟐ

Solution— (a) Given that 푧 = (푐표푠휃 + 푖푠푖푛휃) . From De Moivre’s Theorem we know that—

(푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃) Hence, (푐표푠휃 + 푖푠푖푛휃) = 푐표푠2휃 + 푖푠푖푛2휃. (b) Let 푧 = (1 + 푖) Now, converting 푧 to polar form— |푧| = √1 + 1 = √2 and 휃 = tan =

Hence, 푧 = √2(푐표푠 + 푖푠푖푛 ) Now, there is given (1 + 푖)

So, (1 + 푖) = √2 푐표푠 + 푖푠푖푛 From De Moivre’s Theorem we know that—

(푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃)

Hence, (1 + 푖) = 2 푐표푠 + 푖푠푖푛 = 2 cos 휋 + + 푖푠푖푛 휋 +

= 2 −cos − 푖푠푖푛 = 2 −√− 푖

√= −4(1 + 푖)

(c)Let 푧 = + √ 푖 = 푐표푠 + 푖푠푖푛

From De Moivre’s Theorem we know that— (푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃)

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Hence, + √ 푖 = 푐표푠 + 푖푠푖푛 = 푐표푠 + 푖푠푖푛

= cos 3휋 + + 푖푠푖푛 3휋 + = −푐표푠 − 푖푠푖푛

= − − √ 푖

(푑) Let 푧 =

= ∗ = ( ) = (푐표푠휃 + 푖푠푖푛휃) From De Moivre’s Theorem we know that—

(푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃)

Hence, = cos (24휃) + 푖푠푖푛(24휃) 2.16 Powers to "풊 ”— Let 푧 = 푖, here, 푟 = 1 & 휃 = tan ( ) = tan ∞ = Hence, 푧 = 푖 can be written in Cartesian form as—

푖 = 푐표푠휋2

+ 푖푠푖푛휋2

Thus, 푖 = 푐표푠 + 푖푠푖푛 From De Moivre’s Theorem we know that—

(푐표푠휃 + 푖푠푖푛휃) = cos(푛휃) + 푖푠푖푛(푛휃) Therefore, 푖 = 푐표푠 + 푖푠푖푛

Note— 푖 = 푖, 푖 = −1, 푖 = −푖, 푖 = 1 Hence, 푖 = 푖, 푖 = −1, 푖 = −푖, 푖 = 1 Where, 푛 is an integer. Example—Solve the following complex numbers— (a) 풊ퟗퟗ (b) 풊ퟐퟔ (c) 풊ퟐퟐퟐ (d)풊ퟓퟕ Solution— (a) Given complex number is 푧 = 푖

It can be written as 푧 = 푖( ∗ ) Hence, 푧 = −푖

(b) Given complex number is 푧 = 푖 It can be written as 푧 = 푖( ∗ ) Hence, 푧 = −1



(c) Given complex number is 푧 = 푖 It can be written as 푧 = 푖( ∗ ) Hence, 푧 = −1

(d) Given complex number is 푧 = 푖 It can be written as 푧 = 푖( ∗ ) Hence, 푧 = 푖

2.17 Laws of algebra of complex numbers— The closure law— The sum of two complex numbers is a complex number, i.e., 푧 + 푧 is a complex number for all complex numbers 푧 and 푧 . The commutative law— For any two complex numbers 푧 and 푧 , 푧 + 푧 = 푧 + 푧 . The associative law— For any three complex numbers 푧 , 푧 , 푧 , (푧 + 푧 ) + 푧 = 푧 + (푧 + 푧 ). The distributive law— For any three complex numbers 푧 , 푧 , 푧 , (a) 푧 (푧 + 푧 ) = 푧 푧 + 푧 푧 (b) (푧 + 푧 )푧 = 푧 푧 + 푧 푧 The existence of additive identity— There exists the complex number 0 + 푖 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number 푧, 푧 + 0 = 푧. 2.18 Properties of complex numbers—

1. 풛 = 풛 2. If 풛 = 풙 + 풊풚, then 풌풛 = 풌풙 + 풊풌풚 3. 풛풛 = |풛|ퟐ 4. 풛ퟏ ± 풛ퟐ = 풛ퟏ ± 풛ퟐ 5. 풛ퟏ풛ퟐ = 풛ퟏ 풛ퟐ

6. 풛ퟏ풛ퟐ

= 풛ퟏ풛ퟐ

, 풛ퟐ ≠ ퟎ



EXERCISE-I (a) Find the modulus of the following complex numbers—

1. 2 + 3푖 2. −1 + 2푖 3. 3 + 푖 4. 3 − 4푖 5. 4 + 3푖

(b) Find the argument of the following complex numbers— 1. −1 + 푖 2. 1 − 푖 3. −1 − 푖 4. √3 + 푖 5. 1 + √3푖 6. 1 − 푖푐표푡휃 7. 푠푖푛휃 + 푖(1 − 푐표푠휃) 8. (1 − 푐표푠휃) + 푖푠푖푛휃 9. 푡푎푛휃 − 푖 10. 1 + 푠푖푛휃 + 푖푐표푠휃

(c) Find the complex conjugate of the following complex numbers— 1. 12 + 3푖 2. √3 + 12푖 3. 4 + √3푖 4. 1 − 푖푐표푡휃 5. 푠푖푛휃 + 푖(1 − 푐표푠휃)



6. (1 − 푐표푠휃) + 푖푠푖푛휃 7. 푡푎푛휃 − 푖 8. 1 + 푠푖푛휃 + 푖푐표푠휃

EXERCISE-II (a) If 푧 = 2 + 3푖 , 푧 = −3 + 푖 & 푧 = 1 + 푖. Evaluate the followings—

1. 푧 + 푧 + 푧 2. 푧 + 2푧 − 푧 3. 푧 푧 4.

5. 푧 푧 (b) If 푧 = 푖 , 푧 = −1 + 푖 & 푧 = 1 + √3푖. Evaluate the argument of the followings—

1. 푧 + 푧 + 푧 2. 푧 + 2푧 − 푧 3. 푧 푧 4.

5. 푧 푧

EXERCISE-III (a) Convert the following complex numbers in polar form—

1. √3 + 푖 2. 3 + 3푖 3. 푖 4. −1 + √3푖 5. 2

(b) Convert the following complex numbers in Euler’s form— 1. √3 − 푖 2. −3 + √3푖 3. 푖 4. −1 + √3푖 5. 1 + 푖

(c) Convert the following complex numbers in Euler’s form—

1.

2. √

+√푖

3. − − √ 푖

4. −1 + √3푖 5. 1 + 푖



EXERCISE-IV (a) Solve the following complex numbers—

1. √

+√푖

2. − + √ 푖

3.

4. 5. ((푐표푠휃 + 푖푠푖푛휃)(푐표푠휃 − 푖푠푖푛휃))

(b) Solve the following Complex numbers— 1. 푖 2. 푖 3. 푖 4. 푖 5. 푖

Crackers Attack LEVEL-I

(a) Find the modulus of the following complex numbers— 1. 1 − 푖푐표푡휃 2. 푠푖푛휃 + 푖(1 − 푐표푠휃) 3. (1 − 푐표푠휃) + 푖푠푖푛휃 4. 푡푎푛휃 − 푖 5. 1 + 푠푖푛휃 + 푖푐표푠휃

(b) Find the argument of the following complex numbers— 1. 1 − 푖푐표푡휃 2. 푠푖푛휃 + 푖(1 − 푐표푠휃) 3. (1 − 푐표푠휃) + 푖푠푖푛휃 4. 푡푎푛휃 − 푖 5. 1 + 푠푖푛휃 + 푖푐표푠휃

LEVEL-II (a) Find the square root of the following complex numbers—

1. 8 − 6푖 2. −15 + 8푖 3. (1 − 푐표푠휃) + 푖푠푖푛휃 4. 푡푎푛휃 − 푖 5. 1 + 푠푖푛휃 + 푖푐표푠휃

Advanced Problems



LEVEL-I 1. Evaluate 푖 . 2. Express 푐표푠푥 and 푠푖푛푥 in terms of Euler’s expressions. 3. Prove that sin 푥 + cos 푥 = 1, by using complex numbers. 4. If 푧 = 푥 + 푖푦, evaluate ln z in terms of "푟" and "휃".

LEVEL-II 1. Evaluate (i.e., find all possible values of ) 1 . 2. Evaluate (1 + 푖)√ . 3. Evaluate √푖 .

Reference— 1. en.wikipedia.org/wiki/Complex_number 2. https://www.khanacademy.org 3. www.stewartcalculus.com 4. www.britannica.com 5. mathworld.wolfram.com 6. www.mathsisfun.com 7. www.purplemath.com 8. www.mathwarehouse.com 9. www.clarku.edu 10. home.scarlet.be

https://www.khanacademy.org/

http://www.stewartcalculus.com/

http://www.britannica.com/

http://www.mathsisfun.com/

http://www.purplemath.com/

http://home.scarlet.be/

Diploma_Semester-II_Advanced Mathematics_Complex number

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