Algebraic Operations on Complex Numbers

Complex numbers is an extremely important chapter of Algebra a sit lays the foundation fro various other chapters too. In this section, our focus would be on algebra of complex numbers which shall include equality of two complex numbers, addition, subtraction, multiplication and division of complex numbers. Finally we shall focus on computation of square root of complex number.

In performing operations with complex numbers we can proceed as in the algebra of real numbers replacing i² by -1 when it occurs.

Let z₁ and z₂ be two complex numbers where z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

Then if both the complex numbers are equal i.e. if z₁ = z₂ then Re(z₁) = Re(z₂) and Im(z₁) = Im(z₂).

i.e. if x₁ + iy₁ = x₂ + iy₂

Then x₁ = x₂ and y₁ = y₂ simultaneously.

Let z₁ and z₂ be two complex numbers, z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

Then the addition of the complex numbers z₁ and z₂  is defined as

z₁ + z₂ = (x₁ + iy₁) + ( x₂ + iy₂) = (x₁ + x₂) + i(y₁ + y₂) ∈ C

Re (z₁ + z₂) = Re (z₁) + Re (z₂) and

Im (z₁ + z₂) = Im (z₁) + Im (z₂)

Let z₁ and z₂ be two complex numbers, z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

Then the subtraction of the complex numbers z₁ and z₂ is defined as

z₁ - z₂ = (x₁ + iy₁) - ( x₂ + iy₂) = (x₁ - x₂) + i(y₁ - y₂) ∈ C

Re (z₁ - z₂) = Re(z₁) - Re(z₂) and

Im (z₁ - z₂) = Im(z₁) - Im(z₂)

Watch the following video for more on addition and subtraction of complex numbers:

Let z₁ and z₂ be two complex numbers, z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

Then the multiplication of the complex numbers z₁ and z₂  is defined as

z₁.z₂ = (x₁ + iy₁).(x₂ + iy₂) = (x₁x₂ - y₁y₂ ) + i(x₁y₂ + x₂y₁) ∈ C

Re(z₁ z₂) = Re(z₁) . Re(z₂) - Im(z₁) . Im(z₂)

Im(z₁ z₂) = Im(z₁) . Re(z₂) + Im(z₂) . Re(z₁)

For a real number λ and a complex number z = x + iy,

λ.z = λ(x + iy) = λx + i λy

The following properties are obvious:

(a) λ (z₁ + z₂) = λz₁ + λz₂

(b) λ₁ (λ₂ z) = (λ₁λ₂)z

(c) (λ₁ + λ₂)z = λ₁z + λ₂z for all z, z₁, z₂ ∈ C and λ, λ₁, λ₂ ∈ R.

Let z₁ and z₂ be two complex numbers, z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂.

Then, z₁ / z₂ = (x₁ + iy₁) /(x₂ + iy₂)

                   = (x₁ + iy₁) /(x₂ + iy₂) x (x₂ - iy₂)/(x₂ - iy₂)

                   = (x₁x₂ - x₁y₂i + y₁x₂i - y₁y₂i²) / (x₂² – y₂²i²)

                   = [(x₁x₂ + y₁y₂) + (y₁x₂ – x₁y₂)i ]/ (x₂² – y₂²i²)

                   = (x₁x₂ + y₁y₂)/ (x₂² + y₂²) + (y₁x₂ – x₁y₂)/ (x₂² + y₂²)i

Example 1:  If z = x + iy, z^1/3 = a – ib and x/a – y/b = k(a² - b²), then find the value of k.

Solution: (x + iy)^1/3 = a – ib

So, x + iy = (a – ib)³

               = (a³ – 3ab²) + i(b³ – 3a²b)

So, x = a³ – 3ab², y = b³ – 3a²b

This gives x/a = a² - 3b² and y/b = b² – 3a².

So, x/a – y/b = a² – 3b² – b² + 3a² = 4(a² - b²)

so, k = 4.

Example 2: Show that the polynomial x^4p + x^4q+1 + x^4r+2 + x^4s+3 is divisible by x³ + x² + x + 1 where p, q, r, s ∈ N.

Solution: Let f(x) = x^4p + x^4q+1 + x^4r+2 + x^4s+3

x³ + x² + x + 1 = (x² + 1)(x + 1) = (x + i)(x – i)(x + 1)

f(i) = i^4p + i^4q+1 + i^4r+2 + i^4s+3 = 1 + i + i² + i³ = 1 + i – 1 – i = 0

f(-i) = (- i)^4p + (-i)^4q+1 + (-i)^4r+2 + (-i)^4s+3 = 1 + (-i) + (-i)² + (-i)³ = 1 – i – 1 + i = 0

f(-1) = (- 1)^4p + (-1)^4q+1 + (-1)^4r+2 + (-1)^4s+3 = 0

Thus, by division theorem f(x) is divisible by x³ + x² + x + 1.

Example 3: If the expression (1 + ir)³ is of the form s(1 + i) for some real ‘s’ where ‘r’ is also real and i = √-1, then the value of ‘r’ can be

(a) cot π/8

(b) sec π

(c) tan π/12

(d) tan 5π/12

Solution: We have (1 + ir)³ = s(1 + i)

1+3ri + 3r²i² + r³i³ = s(1 + i)    

1-3r² + i(3r – r³) = s + si

So, 1 – 3r² = s = 3r – r³

Hence, 1 – 3r² = 3r – r³

r³ – 3r² – 3r + 1 = 0 gives (r³ + 1) – 3r(r + 1) = 0

We have (r + 1)(r² + 1 – r – 3r) = 0

Therefore, r = -1 or r² – 4r + 1 = 0

So, r = [4 ± √16 – 4] / 2 = [4 ± 2√3] / 2

So, r = 2 + √3 or 2 - √3.

Hence, options (b), (c) and (d) are correct.

Q1. √a√b = √ab only if

(a) at least one of a or b is non-negative

(b) both a and b are non-negative

(c) at least one of a or b is positive

(d) none of these

Q2. z₁/z₂ =

(a) Re(z₁) + Re(z₂) + i {Im(z₁) + Im (z₂)}

(b) Re(z₁/z₂) + i Im(z₁/z₂)

(c) Re (z₁) Im (z₂) + i Im(z₁) Re(z₂)

(d) none of these

Q3. Re(z₁z₂) =

(a) Re(z₁) + Re(z₂) + i {Im(z₁) + Im (z₂)}

(b) Re(z₁/z₂) + i Im(z₁/z₂)

(c) Re (z₁) .Re (z₂) - Im(z₁) Im(z₂)

(d) none of these

Q4. In context of complex numbers

(a) we can talk of only equality of complex numbers

(b) there is no such concept as equality of complex numbers

(c) we can talk of inequality of complex numbers

(d) we can talk of both equality and inequality of two complex numbers

Q5. The real values of x and y for which the equation (x⁴ + 2xi ) – (3x² + yi) = ( 3 – 5i) + (1 + 2yi) is satisfied.

(a) x = 2 and y = 3.

(b) x = – 2 and y = 1/3.

(c ) x = ± 2 and y = ± 3.

(d) x = ± 2, y = 3, 1/3

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