Chapter-5. Complex Number and Quadratic Equations Mathematics class 11 in english Medium CBSE Notes

7. Algebra of Complex Numbers

(A)   Addition of Complex numbers

Let z₁ = a + ib and z₂ = c + id be any two complex numbers. Then, the sum z₁ + z₂ is defined as follows:

z₁ + z₂ = (a+ib)+(c+id) = (a + c) + i (b + d), which is again a complex number.

Ex: z₁ =2 + i3 and z₂ = 7 + i5

z₁+ z₂ = (2+7) +i(3+5)

           = 9+i8

Alternative Method:

z₁+ z₂ = 2+i3 +7+i5

          = 2+7+i3+i5

          = 9+i(3+5)

          = 9+i8

(B) Subtraction of Complex Numbers

Given any two complex numbers z₁ and z₂, the difference z₁ – z₂ is defined as follows:

z₁ – z₂ = (a+ib) – (c+id) = (a–c)+i(b–d)

For example: z₁= 4+i3,  z₂ =3 + i7

z₁ – z₂ =(4–3)+i(3–7)= 1–i4                  

Note: {4 and 3are like term and i3 and i7 are another like term}

Alternative Method:

z₁ – z₂ =(4+i3) – (3+i7)

            =4+i3 – 3 – i7

            =4– 3+i3 – i7

            =1+i(3 – 7)

            =1-i4

(C) Multiplication of Complex numbers:

z₁ × z₂ = (a +ib) (c+id)

           =a(c+id) +ib(c+id)

           = ac + iad + ibc +( –bd)

           = ac –bd +i(ad+bc)

For example: z­₁= 2+3i, z₂=5 +2i

z­₁× z₂ = (2+3i)( 5 +2i)

           = (2×5 – 3×2)+ i(2×2+3×5)

            = 10 – 6 + i(4+15)

            = 4 + 19i

Alternative Method:

z­₁× z₂ = (2+3i)( 5 +2i)

           =2( 5 +2i) +3i( 5 +2i)

           = 10 + 4i + 15i + 6i²

           = 10 + 4i + 15i + 6(–1)     

           = 10 – 6 + 19i

           = 4 + 19i

(D)Some Important identities:

1. (z₁ + z₂)² = z₁² + 2 z₁ z₂ + z₂²
2. (z₁ – z₂)² = z₁² – 2 z₁ z₂ + z₂²
3. (z₁ + z₂)³ = z₁³ + 3 z₁² z₂ + 3 z₁z₂² + z₂³
4. (z₁ – z₂)³ = z₁³ – 3 z₁² z₂ + 3 z₁z₂² – z₂³
5. z₁² – z₂² = (z₁ + z₂)( z₁ – z₂)