Complex Numbers and Quadratic Equations Class 11 Notes Maths Chapter 5 | Quick Revision Notes-EduGrown Maths Notes

Complex Number

Complex number is of the form a +ib where a is real part and b is imaginary part. Here i = √ -1

 E.g.: 2+ i3 ;  7+ i9 etc

Complex Numbers are used in many scientific fields.

 Two complex numbers are equal if:

• Real parts are equal
• Imaginary parts are equal

E.g. Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.

Algebra of a Complex number

Addition of two complex numbers

Let z1 = a + ib and z2 = c + id be any two complex numbers.   Then, z1 + z2 = (a + c) + i (b + d)

 For example, (2 + i3) + (4 +i5) = 6 + i8

The addition of complex numbers satisfies the following properties:

• Closure law : z1 + z2  = complex Number
• Commutative law: z1 + z2 = z2 + z1
• Associative law: (z1 + z2) + z3 = z1 + (z2 + z3).
• Additive identity : z + 0 = z.
• Additive inverse : z + (–z) = 0.

Difference of two complex numbers

Let z1 = a + ib and z2 = c + id be any two complex numbers.   Then, z1 – z2 = (a – c) + i (b – d)

 For example, (6 + i3) – (2 + i) = 4 + i2

Multiplication of two complex numbers

Let z1 = a + ib and z2 = c + id be any two complex numbers.   Then, z1 * z2 = (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 satisfies the following properties:

• Closure law : z1 * z2  = complex Number
• Commutative law: z1 * z2 = z2 * z1
• Associative law: (z1 * z2) *z3 = z1 * (z2 * z3).
• Multiplicative identity : z * 1 = z.
• Multiplicative inverse : z * (1/z) = 1.    (where z ≠ 0)
• Distributive law  :  z1 (z2 + z3) = z1 z2 + z1 z3

Division of two complex numbers

Given any two complex numbers z1 and z2, where z2 ≠ 0 ,   z1/z2  = z1 * (1/z2)

 For example, let z1 = 2+ 3i and z2 = 2 +2i,

z1* z2 =  (2+ 3i)/ (2+ 2i)

To solve this, we will rationalize the denominator

z1* z2 =  (2+ 3i)/ (2+ 2i)   *  (2- 2i)/ (2- 2i)     =  (-2 + i10) / 8   = -1/4 + i5/4

Power of I

• i² = -1
• i³ = -i
• i⁴ = 1
• i⁵ = i
• i⁶ = -1   etc
• i^-1 = -i
• i^-2 = -1
• i^-3 = i
• i^-4 = 1

Identities

• (z₁ + z₂)² = z₁² + z₂² + 2z₁z₂
• (z₁ – z₂)² = z₁² + z₂² – 2z₁z₂
• (z₁ + z₂)³ = z₁³ + z₂³ + 3z₁z₂² + 3z₁²z₂
• (z₁ – z₂)³ = z₁³ – z₂³ + 3z₁z₂² _– 3z₁²z₂
• z₁² – z₂²  = (z₁ + z₂) (z₁ – z₂)

Refer ExamFear video lessons for Proofs for these identities.

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

Solution:  (5 – 3i)³ = 5³ – 3 × 52 × (3i) + 3 × 5 (3i)² – (3i)³ = 125 – 225i – 135 + 27i = – 10 – 198i.

Modulus & Conjugate  of a complex Number

Let z = a + ib be a complex number.   Modulus of z, denoted by | z |, is defined to be real number (a² + b² )^1/2 ,  | z | = (a² + b²)^1/2

Numerical: Find the Modulus of (3 – 4i )

Solution:  | z | = (a² + b² )^1/2 = (3² + 4²)^1/2  = 5

Let z = a + ib be a complex number.  The conjugate of z, denoted as �, is the complex number a – ib, i.e., �  = a – ib.

Also Z* � = | Z |²

Or   Z^–1 =   � / | Z |²    ( Useful to find inverse of a complex number)

Numerical: Find the conjugate  of  (3 + 4i )

Solution:  Conjugate � = 3-4i

Numerical: Find inverse of  (3 + 4i )

  Z^–1 =   � / | Z |²    = (3 – 4i)/5     = 3/5 – 4/5i

Argand Plane & Polar representation

Complex numbers can represented in 2 forms

• Argand Plane
• Polar Representation

Argand Plane

The complex number x + iy  can be represented  geometrically as the unique point  P(x, y) in the XY-plane and vice-versa. Plane with complex number assigned to each of its point is called complex or Argand plane.

Let’s plot some points on the graph.

Note: Modulus of the complex number is distance between point P(x, y) to the origin O (0, 0)

Polar representation

Let point P represent z = x + iy.  

Let   x = r cos θ , y = r sin θ and therefore, z = r (cos θ + i sin θ).

 Here – π < θ ≤ π

Point P is uniquely determined by the ordered pair of real numbers (r, θ), called the polar coordinates of the point P.

Numerical: Represent the complex number z =1+ i √3 in the polar form.

Solution:  let z =1+ i √3  = r(cos θ + i sin θ)

 r=| z | = (a² + b² )^1/2      = ((1)² + (√3)²)^1/2     = 2

Comparing real parts of  z =1+ i √3  = r(cos θ + i sin θ)   = 2(cos θ + i sin θ) 

1 = 2 cos θ  

or  cos θ   = ½  

or cos θ    = π/3

Therefore,  polar representation will be  z = r(cos θ + i sin θ) = 2(cos π/3 + i sin π/3)

Quadratic Equation

We have seen of real numbers in the cases where discriminant is non-negative, i.e., ≥ 0,

Let us consider the following quadratic equation: ax² + bx + c = 0 with real coefficients a, b, c and a ≠ 0.

Also, let us assume that the b² – 4ac < 0.

Numerical: Solve x² + x + 1= 0

Solution:  Determinant,  b² – 4ac = 12 – 4 × 1 × 1 = 1 – 4 = – 3

X = (-1 ± I √3)/2