Complex Numbers and Quadratic Equations — Class 11 Maths Notes

Topics covered in Complex Numbers and Quadratic Equations

1. 1.Introduction of Complex Numbers and Quadratic Equations

   Complex Numbers
   

   A complex number is a number that can be expressed in the form 𝑎+𝑏𝑖a+bi, where:

   • 𝑎a and 𝑏b are real numbers.
   • 𝑖i is an imaginary unit, defined by the property 𝑖2=−1i2=−1.

   For example, 3+4𝑖3+4i is a complex number where 33 is the real part and 4𝑖4i is the imaginary part.

   Real-life Example:

   • In electrical engineering, complex numbers are used to analyze AC circuits. The real part represents the resistance, and the imaginary part represents the reactance.

   Parts of a Complex Number:

   • Real Part (𝑎a): The real number component.
   • Imaginary Part (𝑏𝑖bi): The imaginary number component.

   Quadratic Equations

   A quadratic equation is a polynomial equation of degree 2, generally written as: 𝑎𝑥2+𝑏𝑥+𝑐=0ax2+bx+c=0 where 𝑎a, 𝑏b, and 𝑐c are constants, and 𝑎≠0a=0.

   Solution of Quadratic Equations:

   • The solutions (roots) of the quadratic equation can be found using the quadratic formula: 𝑥=−𝑏±𝑏2−4𝑎𝑐2𝑎x=2a−b±b2−4ac​​

   The expression under the square root, 𝑏2−4𝑎𝑐b2−4ac, is called the discriminant.

   • If 𝑏2−4𝑎𝑐>0b2−4ac>0: The equation has two distinct real roots.
   • If 𝑏2−4𝑎𝑐=0b2−4ac=0: The equation has exactly one real root (repeated root).
   • If 𝑏2−4𝑎𝑐<0b2−4ac<0: The equation has two complex roots.

   Real-life Example:

   • Quadratic equations are used in physics to describe the motion of objects under the force of gravity. For example, the trajectory of a ball thrown in the air follows a parabolic path, which can be described by a quadratic equation.

   Step-by-Step Example

   Let's solve the quadratic equation 𝑥2−4𝑥+5=0x2−4x+5=0.

   1. Identify 𝑎a, 𝑏b, and 𝑐c:

      • 𝑎=1a=1
      • 𝑏=−4b=−4
      • 𝑐=5c=5

   2. Calculate the discriminant: 𝑏2−4𝑎𝑐=(−4)2−4(1)(5)=16−20=−4b2−4ac=(−4)2−4(1)(5)=16−20=−4

      Since the discriminant is negative, the quadratic equation has two complex roots.

   3. Use the quadratic formula to find the roots: 𝑥=−(−4)±−42(1)=4±−42=4±2𝑖2=2±𝑖x=2(1)−(−4)±−4​​=24±−4​​=24±2i​=2±i

      So, the roots are 2+𝑖2+i and 2−𝑖2−i.

2. 2.Complex Numbers

   Brief Explanation

   Imagine you're playing a game on a grid where you can move left, right, up, and down. Regular numbers, like 2 or -3, only let you move left and right on the grid. But what if you wanted to move diagonally or in other directions? That's where complex numbers come in! They help you move in all directions by combining two types of movements: left-right (real part) and up-down (imaginary part).

   Detailed Explanation

   Real-Life Example

   Think of a complex number as a coordinate on a map. For example, when you use Google Maps, you get directions with coordinates like (3, 4). Here, the number 3 can represent moving right 3 units, and 4 can represent moving up 4 units. Together, these coordinates help you find a specific location.

   In math, we can write these movements using complex numbers. A complex number is written as 𝑎+𝑏𝑖a+bi, where:

   • 𝑎a is the real part (left-right movement).
   • 𝑏b is the imaginary part (up-down movement).
   • 𝑖i is an imaginary unit, which is defined as 𝑖2=−1i2=−1.

   Understanding Complex Numbers

   1. Imaginary Unit 𝑖i:

      • The imaginary unit 𝑖i is a special number such that 𝑖2=−1i2=−1. This means 𝑖i is not a real number but helps in extending our number system.

   2. Complex Number 𝑎+𝑏𝑖a+bi:

      • Any complex number is made up of a real part 𝑎a and an imaginary part 𝑏b.

   Example:

   Let's take the complex number 3+4𝑖3+4i:

   • 33 is the real part.
   • 4𝑖4i is the imaginary part.

   This means you move 3 units to the right (real part) and 4 units up (imaginary part) on our grid.

   Operations with Complex Numbers

   1. Addition:

      • To add two complex numbers, add their real parts and their imaginary parts separately.
      • Example: (3+4𝑖)+(1+2𝑖)=(3+1)+(4𝑖+2𝑖)=4+6𝑖(3+4i)+(1+2i)=(3+1)+(4i+2i)=4+6i.

   2. Subtraction:

      • Subtract the real parts and the imaginary parts separately.
      • Example: (3+4𝑖)−(1+2𝑖)=(3−1)+(4𝑖−2𝑖)=2+2𝑖(3+4i)−(1+2i)=(3−1)+(4i−2i)=2+2i.

   3. Multiplication:

      • Use the distributive property and remember that 𝑖2=−1i2=−1.
      • Example: (3+4𝑖)×(1+2𝑖)(3+4i)×(1+2i).
      • Calculation: 3×1+3×2𝑖+4𝑖×1+4𝑖×2𝑖3×1+3×2i+4i×1+4i×2i
      • Simplify: 3+6𝑖+4𝑖+8𝑖2=3+10𝑖+8(−1)=3+10𝑖−8=−5+10𝑖3+6i+4i+8i2=3+10i+8(−1)=3+10i−8=−5+10i.

   4. Division:

      • Divide two complex numbers by multiplying the numerator and the denominator by the conjugate of the denominator.
      • Example: 3+4𝑖1+2𝑖1+2i3+4i​.
      • Conjugate of 1+2𝑖1+2i is 1−2𝑖1−2i.
      • Calculation: (3+4𝑖)×(1−2𝑖)(1+2𝑖)×(1−2𝑖)=(3+4𝑖)×(1−2𝑖)12−(2𝑖)2(1+2i)×(1−2i)(3+4i)×(1−2i)​=12−(2i)2(3+4i)×(1−2i)​
      • Simplify: 3−6𝑖+4𝑖−8𝑖21−4(−1)=3−2𝑖+81+4=11−2𝑖5=115−2𝑖5=2.2−0.4𝑖1−4(−1)3−6i+4i−8i2​=1+43−2i+8​=511−2i​=511​−52i​=2.2−0.4i.

   Applications in Real Life

   1. Engineering: Complex numbers are used in electrical engineering to analyze AC circuits.
   2. Physics: They help in solving problems related to wave functions in quantum mechanics.
   3. Computer Graphics: Complex numbers are used in generating fractals and in signal processing.
   4. Control Systems: They help in designing and analyzing control systems for stability and response.

3. 3.Algebra of Complex Numbers

   Complex numbers extend the idea of one-dimensional real numbers to the two-dimensional complex plane by introducing an imaginary unit
   

   𝑖i where 𝑖2=−1i2=−1. Let's explore how to perform algebraic operations with complex numbers.

   Brief Explanation

   Complex numbers are written in the form 𝑎+𝑏𝑖a+bi, where 𝑎a is the real part and 𝑏b is the imaginary part. The algebra of complex numbers involves basic operations like addition, subtraction, multiplication, and division. Understanding these operations helps in solving complex equations and applying these concepts in various fields like engineering and physics.

   Detailed Explanation

   1. Addition

   To add two complex numbers, add their real parts and their imaginary parts separately.

   Formula: (𝑎+𝑏𝑖)+(𝑐+𝑑𝑖)=(𝑎+𝑐)+(𝑏+𝑑)𝑖(a+bi)+(c+di)=(a+c)+(b+d)i

   Example: (3+4𝑖)+(1+2𝑖)=(3+1)+(4+2)𝑖=4+6𝑖(3+4i)+(1+2i)=(3+1)+(4+2)i=4+6i

   2. Subtraction

   To subtract two complex numbers, subtract their real parts and their imaginary parts separately.

   Formula: (𝑎+𝑏𝑖)−(𝑐+𝑑𝑖)=(𝑎−𝑐)+(𝑏−𝑑)𝑖(a+bi)−(c+di)=(a−c)+(b−d)i

   Example: (3+4𝑖)−(1+2𝑖)=(3−1)+(4−2)𝑖=2+2𝑖(3+4i)−(1+2i)=(3−1)+(4−2)i=2+2i

   3. Multiplication

   To multiply two complex numbers, use the distributive property and the fact that 𝑖2=−1i2=−1.

   Formula: (𝑎+𝑏𝑖)×(𝑐+𝑑𝑖)=(𝑎𝑐−𝑏𝑑)+(𝑎𝑑+𝑏𝑐)𝑖(a+bi)×(c+di)=(ac−bd)+(ad+bc)i

   Example: (3+4𝑖)×(1+2𝑖)=3×1+3×2𝑖+4𝑖×1+4𝑖×2𝑖(3+4i)×(1+2i)=3×1+3×2i+4i×1+4i×2i =3+6𝑖+4𝑖+8𝑖2=3+6i+4i+8i2 =3+10𝑖+8(−1)=3+10i+8(−1) =3+10𝑖−8=3+10i−8 =−5+10𝑖=−5+10i

   4. Division

   To divide two complex numbers, multiply the numerator and the denominator by the conjugate of the denominator and simplify.

   Formula: 𝑎+𝑏𝑖𝑐+𝑑𝑖=(𝑎+𝑏𝑖)×(𝑐−𝑑𝑖)(𝑐+𝑑𝑖)×(𝑐−𝑑𝑖)c+dia+bi​=(c+di)×(c−di)(a+bi)×(c−di)​ =(𝑎+𝑏𝑖)×(𝑐−𝑑𝑖)𝑐2+𝑑2=c2+d2(a+bi)×(c−di)​

   Example: 3+4𝑖1+2𝑖1+2i3+4i​

   • Conjugate of 1+2𝑖1+2i is 1−2𝑖1−2i.
   • Multiply numerator and denominator by the conjugate: (3+4𝑖)×(1−2𝑖)(1+2𝑖)×(1−2𝑖)=(3+4𝑖)×(1−2𝑖)12−(2𝑖)2(1+2i)×(1−2i)(3+4i)×(1−2i)​=12−(2i)2(3+4i)×(1−2i)​
   • Simplify: =(3−6𝑖+4𝑖−8𝑖2)1+4=3−2𝑖+85=11−2𝑖5=2.2−0.4𝑖=1+4(3−6i+4i−8i2)​=53−2i+8​=511−2i​=2.2−0.4i

   Complex Conjugate

   The conjugate of a complex number 𝑎+𝑏𝑖a+bi is 𝑎−𝑏𝑖a−bi. It is useful in division and in finding the magnitude of a complex number.

   Example: Conjugate of 3+4𝑖3+4i is 3−4𝑖3−4i.

   Magnitude (Modulus)

   The magnitude (or modulus) of a complex number 𝑎+𝑏𝑖a+bi is given by: ∣𝑎+𝑏𝑖∣=𝑎2+𝑏2∣a+bi∣=a2+b2​

   Example: Magnitude of 3+4𝑖3+4i: ∣3+4𝑖∣=32+42=9+16=25=5∣3+4i∣=32+42​=9+16​=25​=5

   Applications in Real Life

   1. Electrical Engineering: Analyzing AC circuits using complex numbers.
   2. Quantum Physics: Wave functions are represented using complex numbers.
   3. Signal Processing: Complex numbers are used in Fourier transforms.
   4. Control Systems: Stability and control analysis often use complex numbers.

4. 4.Operations on Complex Numbers

   1. Addition of Two Complex Numbers

   When adding two complex numbers, you simply add their real parts and their imaginary parts separately.

   Formula: (𝑎+𝑏𝑖)+(𝑐+𝑑𝑖)=(𝑎+𝑐)+(𝑏+𝑑)𝑖(a+bi)+(c+di)=(a+c)+(b+d)i

   Example: Consider the complex numbers 3+4𝑖3+4i and 1+2𝑖1+2i: (3+4𝑖)+(1+2𝑖)=(3+1)+(4+2)𝑖(3+4i)+(1+2i)=(3+1)+(4+2)i =4+6𝑖=4+6i

   2. Difference of Two Complex Numbers

   When subtracting two complex numbers, you subtract their real parts and their imaginary parts separately.

   Formula: (𝑎+𝑏𝑖)−(𝑐+𝑑𝑖)=(𝑎−𝑐)+(𝑏−𝑑)𝑖(a+bi)−(c+di)=(a−c)+(b−d)i

   Example: Consider the complex numbers 3+4𝑖3+4i and 1+2𝑖1+2i: (3+4𝑖)−(1+2𝑖)=(3−1)+(4−2)𝑖(3+4i)−(1+2i)=(3−1)+(4−2)i =2+2𝑖=2+2i

   3. Multiplication of Two Complex Numbers

   When multiplying two complex numbers, use the distributive property (also known as the FOIL method for binomials) and remember that 𝑖2=−1i2=−1.

   Formula: (𝑎+𝑏𝑖)×(𝑐+𝑑𝑖)=(𝑎𝑐−𝑏𝑑)+(𝑎𝑑+𝑏𝑐)𝑖(a+bi)×(c+di)=(ac−bd)+(ad+bc)i

   Example: Consider the complex numbers 3+4𝑖3+4i and 1+2𝑖1+2i: (3+4𝑖)×(1+2𝑖)(3+4i)×(1+2i) First, apply the distributive property: =3×1+3×2𝑖+4𝑖×1+4𝑖×2𝑖=3×1+3×2i+4i×1+4i×2i =3+6𝑖+4𝑖+8𝑖2=3+6i+4i+8i2 Since 𝑖2=−1i2=−1: =3+10𝑖+8(−1)=3+10i+8(−1) =3+10𝑖−8=3+10i−8 =−5+10𝑖=−5+10i

   4. Division of Two Complex Numbers

   To divide two complex numbers, multiply the numerator and the denominator by the conjugate of the denominator and simplify.

   Formula: 𝑎+𝑏𝑖𝑐+𝑑𝑖=(𝑎+𝑏𝑖)×(𝑐−𝑑𝑖)(𝑐+𝑑𝑖)×(𝑐−𝑑𝑖)=(𝑎+𝑏𝑖)×(𝑐−𝑑𝑖)𝑐2+𝑑2c+dia+bi​=(c+di)×(c−di)(a+bi)×(c−di)​=c2+d2(a+bi)×(c−di)​

   Example: Consider the complex numbers 3+4𝑖3+4i and 1+2𝑖1+2i: 3+4𝑖1+2𝑖1+2i3+4i​ The conjugate of 1+2𝑖1+2i is 1−2𝑖1−2i. Multiply both the numerator and the denominator by this conjugate: (3+4𝑖)×(1−2𝑖)(1+2𝑖)×(1−2𝑖)(1+2i)×(1−2i)(3+4i)×(1−2i)​ First, multiply the numerator: =(3+4𝑖)×(1−2𝑖)=(3+4i)×(1−2i) =3×1−3×2𝑖+4𝑖×1−4𝑖×2𝑖=3×1−3×2i+4i×1−4i×2i =3−6𝑖+4𝑖−8𝑖2=3−6i+4i−8i2 =3−6𝑖+4𝑖−8(−1)=3−6i+4i−8(−1) =3−6𝑖+4𝑖+8=3−6i+4i+8 =11−2𝑖=11−2i Now, multiply the denominator: (1+2𝑖)×(1−2𝑖)(1+2i)×(1−2i) =1×1−1×2𝑖+2𝑖×1−2𝑖×2𝑖=1×1−1×2i+2i×1−2i×2i =1−2𝑖+2𝑖−4𝑖2=1−2i+2i−4i2 =1−4(−1)=1−4(−1) =1+4=1+4 =5=5 Thus, the division is: 11−2𝑖5511−2i​ This can be separated into real and imaginary parts: =115−2𝑖5=511​−52i​ =2.2−0.4𝑖=2.2−0.4i

   5. Powers of 𝑖i

   The imaginary unit 𝑖i has a cyclical pattern when raised to different powers. Knowing this pattern helps in simplifying expressions involving powers of 𝑖i.

   • 𝑖1=𝑖i1=i
   • 𝑖2=−1i2=−1
   • 𝑖3=−𝑖i3=−i
   • 𝑖4=1i4=1
   • 𝑖5=𝑖i5=i (because 𝑖5=𝑖4+1=𝑖4×𝑖=1×𝑖=𝑖i5=i4+1=i4×i=1×i=i)

   This pattern repeats every four powers.

   Example: Let's find 𝑖17i17: Since the powers of 𝑖i repeat every four numbers, find the remainder when 17 is divided by 4: 17÷4=4 remainder 117÷4=4 remainder 1 Thus, 𝑖17=𝑖1=𝑖i17=i1=i

5. 5.Modulus and Conjugate of a Complex Number

   Brief Explanation:
   

   1. Modulus of a Complex Number:

      • The modulus (or absolute value) of a complex number 𝑧=𝑎+𝑏𝑖z=a+bi is given by ∣𝑧∣∣z∣.
      • It is calculated as ∣𝑧∣=𝑎2+𝑏2∣z∣=a2+b2​, where 𝑎a is the real part and 𝑏b is the imaginary part of the complex number.

   2. Conjugate of a Complex Number:

      • The conjugate of a complex number 𝑧=𝑎+𝑏𝑖z=a+bi is denoted by 𝑧‾z.
      • It is obtained by changing the sign of the imaginary part: 𝑧‾=𝑎−𝑏𝑖z=a−bi.

   Detailed Explanation with Real-Life Example

   1. Modulus of a Complex Number:

   Let's consider a complex number 𝑧=3+4𝑖z=3+4i.

   • Here, 𝑎=3a=3 and 𝑏=4b=4.
   • The modulus is calculated as:∣𝑧∣=𝑎2+𝑏2=32+42=9+16=25=5∣z∣=a2+b2​=32+42​=9+16​=25​=5

   Example: Distance in a Plane

   Think of a complex number as a point in a 2D plane where the x-axis represents the real part and the y-axis represents the imaginary part. The modulus represents the distance from the origin (0,0) to this point. For 𝑧=3+4𝑖z=3+4i, the point is (3, 4) and the distance to the origin is 5 units.

   2. Conjugate of a Complex Number:

   For the same complex number 𝑧=3+4𝑖z=3+4i,

   • The conjugate is:𝑧‾=3−4𝑖z=3−4i

   Example: Electrical Engineering

   In electrical engineering, complex numbers are used to represent impedances. The conjugate of an impedance is used in calculations involving power. If 𝑧z represents an impedance, then 𝑧‾z is used to find the actual power consumed in the circuit.

   Activity

   To practice, find the modulus and conjugate of the following complex numbers:

   1. 𝑧1=1+2𝑖z1​=1+2i
   2. 𝑧2=−3−4𝑖z2​=−3−4i

   Solution:

   1. For 𝑧1=1+2𝑖z1​=1+2i:

      • Modulus: ∣𝑧1∣=12+22=1+4=5∣z1​∣=12+22​=1+4​=5​
      • Conjugate: 𝑧1‾=1−2𝑖z1​​=1−2i

   2. For 𝑧2=−3−4𝑖z2​=−3−4i:

      • Modulus: ∣𝑧2∣=(−3)2+(−4)2=9+16=25=5∣z2​∣=(−3)2+(−4)2​=9+16​=25​=5
      • Conjugate: 𝑧2‾=−3+4𝑖z2​​=−3+4i

   Careers and Real-Life Applications

   • Engineering: Complex numbers are essential in electrical and mechanical engineering for analyzing circuits and systems.
   • Physics: They are used in quantum mechanics and wave theory.
   • Computer Graphics: Complex numbers help in transformations and rotations.
   • Signal Processing: Modulus and conjugate are used in filtering and analyzing signals.

6. 6.Exercise Questions

   Express each of the complex numbers given in Exercises 1 to 6 in the form 𝑎+𝑖𝑏a+ib.

   1. (5𝑖)(35−𝑖)(5i)(53​−i)

   Solution:

   First, we distribute 5𝑖5i across the terms in the parentheses:

   5𝑖(35−𝑖)=5𝑖⋅35−5𝑖⋅𝑖5i(53​−i)=5i⋅53​−5i⋅i

   Calculate each term separately:

   1. 5𝑖⋅35=3𝑖5i⋅53​=3i
   2. 5𝑖⋅𝑖=5𝑖2=5(−1)=−55i⋅i=5i2=5(−1)=−5

   Now, combine the results:

   3𝑖−53i−5

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   −5+3𝑖−5+3i

   2. 𝑖9+𝑖19i9+i19

   Solution:

   We know that 𝑖i has a cycle of 4: 𝑖,−1,−𝑖,1i,−1,−i,1.

   • 𝑖9=𝑖4⋅2+1=𝑖1=𝑖i9=i4⋅2+1=i1=i
   • 𝑖19=𝑖4⋅4+3=𝑖3=−𝑖i19=i4⋅4+3=i3=−i

   Now, add the results:

   𝑖+(−𝑖)=0i+(−i)=0

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   0+0𝑖0+0i

   3. 𝑖−39i−39

   Solution:

   We know that 𝑖i has a cycle of 4: 𝑖,−1,−𝑖,1i,−1,−i,1.

   • 𝑖−39=𝑖−(4⋅9+3)=𝑖−3=1𝑖3=1−𝑖=𝑖−𝑖2=𝑖1=−𝑖i−39=i−(4⋅9+3)=i−3=i31​=−i1​=−i2i​=1i​=−i

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   0−𝑖0−i

   4. (15+𝑖25)−(4+𝑖52)(51​+i52​)−(4+i25​)

   Solution:

   Subtract the real parts and the imaginary parts separately:

   Real part: 15−4=15−205=−19551​−4=51​−520​=−519​

   Imaginary part: 25−52=25−2510=410−2510=−2110=−2.1152​−25​=52​−1025​=104​−1025​=−1021​=−12.1​

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   −195−2.1𝑖−519​−2.1i

   5. [(13+𝑖73)+(4+𝑖13)]−(−43+𝑖)[(31​+i37​)+(4+i31​)]−(−34​+i)

   Solution:

   First, add the terms inside the brackets:

   Real part: 13+4=13+123=13331​+4=31​+312​=313​

   Imaginary part: 73+13=8337​+31​=38​

   Now, subtract the given term:

   Real part: 133−(−43)=133+43=173313​−(−34​)=313​+34​=317​

   Imaginary part: 83−𝑖=83−𝑖38​−i=38​−i

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   173+53𝑖317​+35​i

   6. (1−𝑖)4(1−i)4

   Solution:

   We can expand (1−𝑖)4(1−i)4 using the binomial theorem, but it's easier to use De Moivre's theorem.

   First, express 1−𝑖1−i in polar form:

   1−𝑖=2(cos⁡(−𝜋4)+𝑖sin⁡(−𝜋4))1−i=2​(cos(−4π​)+isin(−4π​))

   Using De Moivre's theorem:

   (1−𝑖)4=(2)4(cos⁡(4×−𝜋4)+𝑖sin⁡(4×−𝜋4))=4(cos⁡(−𝜋)+𝑖sin⁡(−𝜋))=4(−1+0𝑖)=−4(1−i)4=(2​)4(cos(4×−4π​)+isin(4×−4π​))=4(cos(−π)+isin(−π))=4(−1+0i)=−4

   So, the answer in the form 𝑎+𝑖𝑏a+ib is:

   −4+0𝑖−4+0i

7. 7.Argand Plane and Polar Representation

   Brief Explanation:

   1. Argand Plane:

      • The Argand plane is a way to represent complex numbers geometrically. It is similar to a coordinate plane, with the x-axis representing the real part and the y-axis representing the imaginary part of a complex number.

   2. Polar Representation:

      • Polar representation expresses a complex number in terms of its magnitude (or modulus) and angle (or argument). A complex number 𝑧=𝑎+𝑏𝑖z=a+bi can be represented as 𝑧=𝑟(cos⁡𝜃+𝑖sin⁡𝜃)z=r(cosθ+isinθ), where 𝑟r is the modulus and 𝜃θ is the argument.

   Here is the graphical representation of the complex number 𝑧=3+4𝑖z=3+4i in both the Argand Plane and Polar form:

   

   
   

   Argand Plane

   In the Argand Plane:

   • The horizontal axis represents the real part of the complex number.
   • The vertical axis represents the imaginary part of the complex number.

   For 𝑧=3+4𝑖z=3+4i:

   • The point (3, 4) is plotted, showing the position of the complex number.

   Polar Representation

   In Polar Representation:

   • The complex number is represented by its modulus (r) and argument (θ).
   • Modulus 𝑟=5r=5 (distance from the origin to the point).
   • Argument 𝜃≈53.13∘θ≈53.13∘ (angle with the positive real axis).

   The polar plot shows the vector representation of the complex number in polar coordinates.

   These visualizations help in understanding how complex numbers can be represented in different ways for easier manipulation and comprehension. ​

   Detailed Explanation with Real-Life Example

   1. Argand Plane:

   The Argand plane is named after the Swiss mathematician Jean-Robert Argand. It is used to visualize complex numbers. On this plane:

   • The horizontal axis (x-axis) represents the real part of the complex number.
   • The vertical axis (y-axis) represents the imaginary part of the complex number.

   For example, consider the complex number 𝑧=3+4𝑖z=3+4i:

   • The real part is 3, and the imaginary part is 4.
   • On the Argand plane, this is represented as the point (3, 4).

   Example:

   Imagine you want to represent the complex number 3+4𝑖3+4i on the Argand plane. You plot a point at coordinates (3, 4). This visual representation helps understand the magnitude and direction of complex numbers.

   2. Polar Representation:

   A complex number 𝑧=𝑎+𝑏𝑖z=a+bi can also be represented in polar form. The polar form uses the distance from the origin (modulus) and the angle with the positive real axis (argument).

   • Modulus 𝑟r: The distance from the origin to the point (a, b) in the Argand plane.𝑟=𝑎2+𝑏2r=a2+b2​
   • Argument 𝜃θ: The angle formed with the positive x-axis.𝜃=tan⁡−1(𝑏𝑎)θ=tan−1(ab​)

   Using these, the polar form of 𝑧=𝑎+𝑏𝑖z=a+bi is:

   𝑧=𝑟(cos⁡𝜃+𝑖sin⁡𝜃)z=r(cosθ+isinθ)

   Example:

   For 𝑧=3+4𝑖z=3+4i:

   • Modulus 𝑟=32+42=9+16=25=5r=32+42​=9+16​=25​=5
   • Argument 𝜃=tan⁡−1(43)≈53.13∘θ=tan−1(34​)≈53.13∘

   So, in polar form:

   𝑧=5(cos⁡53.13∘+𝑖sin⁡53.13∘)z=5(cos53.13∘+isin53.13∘)

   Activity

   To practice, convert the following complex numbers into polar form and plot them on the Argand plane:

   1. 𝑧1=1+𝑖z1​=1+i
   2. 𝑧2=−2+2𝑖z2​=−2+2i

   Solution:

   1. For 𝑧1=1+𝑖z1​=1+i:

      • Modulus 𝑟=12+12=2r=12+12​=2​
      • Argument 𝜃=tan⁡−1(1)=45∘θ=tan−1(1)=45∘
      • Polar form: 𝑧1=2(cos⁡45∘+𝑖sin⁡45∘)z1​=2​(cos45∘+isin45∘)

   2. For 𝑧2=−2+2𝑖z2​=−2+2i:

      • Modulus 𝑟=(−2)2+22=4+4=8=22r=(−2)2+22​=4+4​=8​=22​
      • Argument 𝜃=tan⁡−1(2−2)=tan⁡−1(−1)=135∘θ=tan−1(−22​)=tan−1(−1)=135∘
      • Polar form: 𝑧2=22(cos⁡135∘+𝑖sin⁡135∘)z2​=22​(cos135∘+isin135∘)

   Careers and Real-Life Applications

   • Engineering: Complex numbers and their representations are crucial in electrical and mechanical engineering for analyzing circuits and mechanical vibrations.
   • Physics: They are used in wave theory, quantum mechanics, and signal processing.
   • Computer Graphics: Complex transformations and rotations often use polar forms.
   • Robotics: Understanding movement and positioning often involves complex numbers and polar coordinates.

8. 8.Exercise Questions

   1. If 𝛼α and 𝛽β are different complex numbers with ∣𝛽∣=1∣β∣=1, then find ∣𝛽−𝛼1−𝛼‾𝛽∣∣∣​1−αββ−α​∣∣​.

   Solution:

   Given ∣𝛽∣=1∣β∣=1, we know that 𝛽𝛽‾=1ββ​=1. This property will be used in our solution.

   The expression to find is ∣𝛽−𝛼1−𝛼‾𝛽∣∣∣​1−αββ−α​∣∣​.

   1. Let's denote 𝑧=∣𝛽−𝛼1−𝛼‾𝛽∣z=∣∣​1−αββ−α​∣∣​.

   2. Using the property ∣𝛽∣=1∣β∣=1, we have:

      ∣𝛽∣=∣𝛽𝛽‾∣=1∣β∣=∣ββ​∣=1

   3. Now consider the modulus property:

      ∣𝛽−𝛼1−𝛼‾𝛽∣=∣𝛽−𝛼∣∣1−𝛼‾𝛽∣∣∣​1−αββ−α​∣∣​=∣1−αβ∣∣β−α∣​

   4. Since ∣𝛽∣=1∣β∣=1, we know 𝛽‾=1𝛽β​=β1​, thus:

      1−𝛼‾𝛽=1−𝛼‾𝛽=𝛽−𝛼1−αβ=1−βα​=β−α

   5. Therefore:

      ∣𝛽−𝛼1−𝛼‾𝛽∣=∣𝛽−𝛼∣∣1−𝛼‾𝛽∣=1∣∣​1−αββ−α​∣∣​=∣1−αβ∣∣β−α∣​=1

   So, the answer is ∣𝛽−𝛼1−𝛼‾𝛽∣=1∣∣​1−αββ−α​∣∣​=1.

   2. Find the number of non-zero integral solutions of the equation ∣1−𝑖∣𝑥=2𝑥∣1−i∣x=2x.

   Solution:

   1. First, calculate ∣1−𝑖∣∣1−i∣:

      ∣1−𝑖∣=12+(−1)2=1+1=2∣1−i∣=12+(−1)2​=1+1​=2​

   2. Substitute ∣1−𝑖∣∣1−i∣ into the equation:

      (2)𝑥=2𝑥(2​)x=2x

   3. Equate the bases:

      (21/2)𝑥=2𝑥(21/2)x=2x

   4. Simplify the exponents:

      2𝑥/2=2𝑥2x/2=2x

   5. Set the exponents equal to each other:

      𝑥2=𝑥2x​=x

   6. Solve for 𝑥x:

      𝑥=0x=0

   Since the question asks for non-zero integral solutions and 𝑥=0x=0 is the only solution, there are no non-zero integral solutions.

   So, the answer is 0 non-zero integral solutions.

   3. If (𝑎+𝑖𝑏)(𝑐+𝑖𝑑)(𝑒+𝑖𝑓)(𝑔+𝑖ℎ)=𝐴+𝑖𝐵(a+ib)(c+id)(e+if)(g+ih)=A+iB, then show that (𝑎2+𝑏2)(𝑐2+𝑑2)(𝑒2+𝑓2)(𝑔2+ℎ2)=𝐴2+𝐵2(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2.

   Solution:

   1. Given: (𝑎+𝑖𝑏)(𝑐+𝑖𝑑)(𝑒+𝑖𝑓)(𝑔+𝑖ℎ)=𝐴+𝑖𝐵(a+ib)(c+id)(e+if)(g+ih)=A+iB.

   2. Let's denote:

      𝑧1=𝑎+𝑖𝑏,𝑧2=𝑐+𝑖𝑑,𝑧3=𝑒+𝑖𝑓,𝑧4=𝑔+𝑖ℎz1​=a+ib,z2​=c+id,z3​=e+if,z4​=g+ih

   3. The product of these complex numbers:

      𝑧1𝑧2𝑧3𝑧4=𝐴+𝑖𝐵z1​z2​z3​z4​=A+iB

   4. Calculate the modulus of each complex number:

      ∣𝑧1∣=𝑎2+𝑏2,∣𝑧2∣=𝑐2+𝑑2,∣𝑧3∣=𝑒2+𝑓2,∣𝑧4∣=𝑔2+ℎ2∣z1​∣=a2+b2​,∣z2​∣=c2+d2​,∣z3​∣=e2+f2​,∣z4​∣=g2+h2​

   5. The modulus of the product is the product of the moduli:

      ∣𝑧1𝑧2𝑧3𝑧4∣=∣𝑧1∣∣𝑧2∣∣𝑧3∣∣𝑧4∣∣z1​z2​z3​z4​∣=∣z1​∣∣z2​∣∣z3​∣∣z4​∣

   6. Substitute the values:

      ∣𝑧1𝑧2𝑧3𝑧4∣=𝑎2+𝑏2⋅𝑐2+𝑑2⋅𝑒2+𝑓2⋅𝑔2+ℎ2∣z1​z2​z3​z4​∣=a2+b2​⋅c2+d2​⋅e2+f2​⋅g2+h2​

   7. Since the modulus of 𝐴+𝑖𝐵A+iB is 𝐴2+𝐵2A2+B2​:

      𝐴2+𝐵2=(𝑎2+𝑏2)(𝑐2+𝑑2)(𝑒2+𝑓2)(𝑔2+ℎ2)A2+B2​=(a2+b2)(c2+d2)(e2+f2)(g2+h2)​

   8. Square both sides to remove the square roots:

      𝐴2+𝐵2=(𝑎2+𝑏2)(𝑐2+𝑑2)(𝑒2+𝑓2)(𝑔2+ℎ2)A2+B2=(a2+b2)(c2+d2)(e2+f2)(g2+h2)

   Thus, we have shown that:

   (𝑎2+𝑏2)(𝑐2+𝑑2)(𝑒2+𝑓2)(𝑔2+ℎ2)=𝐴2+𝐵2(a2+b2)(c2+d2)(e2+f2)(g2+h2)=A2+B2

   4. If (1+𝑖1−𝑖)𝑚=1(1−i1+i​)m=1, then find the least positive integral value of 𝑚m.

   Solution:

   1. Simplify the expression 1+𝑖1−𝑖1−i1+i​:

      1+𝑖1−𝑖⋅1+𝑖1+𝑖=(1+𝑖)2(1−𝑖)(1+𝑖)=1+2𝑖+𝑖21−𝑖2=1+2𝑖−11+1=2𝑖2=𝑖1−i1+i​⋅1+i1+i​=(1−i)(1+i)(1+i)2​=1−i21+2i+i2​=1+11+2i−1​=22i​=i

   2. The equation becomes 𝑖𝑚=1im=1.

   3. The powers of 𝑖i cycle every 4: 𝑖,−1,−𝑖,1i,−1,−i,1.

   4. For 𝑖𝑚=1im=1, 𝑚m must be a multiple of 4.

   5. The smallest positive integer value of 𝑚m is 4.

   So, the least positive integral value of 𝑚m is 44​.