Consider two complex numbers

 z₁ = a₁+ ib₁

 z₂ = a₁ + ib₂

z₁∗ z₂= (a₁+i b₁) (a₂ +ib₂)

     = a₁ (a₂ +ib₂) + i b₁(a₂ + ib₂)

     = a₁ a₂ + i a₁b₂ + i a₂ b₁ + i₂ b₁b₂

     = (a₁ a₂ - b₁b₂ )+ i(a₁b₂ + a₂b₁)

Since i₂ =-1

 i₂ b₁b₂ = - b₁b₂

Multiply the following complex numbers

Example 1

(2+ 3i)∗ (4-5i)

Let z₁ =2+ 3i,    a₁ = 2,      b₁ = 3

      z₂ = 4-5i,     a₂ = 4,      b₂ = -5

Comparing with z₁ z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂ + a₂ b₁)

(2 + 3i) (4 - 5i) = (2 ∗4 - 3 ∗(-5) + i (2 ∗(-5) + 4 ∗3)

(8+15) +i (-10+ 12)

                     = 23 + 2i

Example 2

Multiply (-6 - 5i) and (-7 + 2i)

Let z₁ = -6 - 5i,    a₁ = -6,       b₁ = -5

     z₂ = -7 + 2i,   a₂ = -7,       b₂ = 2

Comparing with z₁ ∗z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)

(-6 - 5i) (-7 + 2i) = [(-6) ∗(-7) - (-5) ∗(2)] + i ((-6)∗ (2) + (-7) ∗(-5))

                       = (42 + 10) + i (-12 + 35)

                       = 52 + 23i

Example 3

Multiply (1/2 - 4/3i) and (2/7 - 4/5i)

Let z₁ =1/2 - 4/3i,      a₁= 1/2,       b₁ = -4/3

     z₂ = 2/7 - 4/5i,    a₂ = 2/7,      b₂ = -4/5

Comparing with z₁ ∗z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)

  (1/2 - 4/3 i) (2/7 - 4/5i) = (1/2 ∗2/7 - (-4/3) (-4/5))

                                            + i( 1/2 ∗(-4/5) +2/7 ∗(-4/3))

                                = (1/7 - 16/15) + i (-2/5 - 8/21)

                                = ((15-112)/105) +i ((-42-40)/105)

                                = -97/105 - 82/105i

Example 4

Multiply (9 - 3i) and (9 + 3i)

Let z₁ = 9 - 3i,     a₁ = 9,    b₁ = -3

     z₂ = 9 + 3i,     a₂ = 9,    b₂ = 3

Note. z₂ = or z₂ is the complex conjugate of z₁

Comparing with z₁ ∗z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)

               (9-3i) ∗(9+3i) = (9∗9 - (-3) (3)) +i (9∗3 + (9) (-3))

                                                     (81+ 9) +i (27- 27)

                                   = 90 + 0i

                                   = 90.

Alternately the above problem can be solved using the difference of squares.

Since (a-b) (a+b) = a² -b²

    (9- 3i) (9+ 3i) = (9)² - (3i)²

                          = 81 - 9i²s

                          = 81 - 9(-1)  since i² = -1

                          = 81 + 9

                          = 90

Example 5

Multiply (-6 + 7i) ∗12i

Let z₁ = -6 + 7i,    a₁ = -6,    b₁ = 7

     z₂ = 12i,         a₂ = 0,      b₂ = 12

Comparing with z₁ ∗z₂ = (a₁ a₂- b₁ b₂) +i(a₁ b₂+ a₂ b₁ )

(-6 + 7i) .12i= (-6) ∗0 - 7∗12) + i( -6∗12+ 0 ∗7)

                 = -84 - 72i

Alternately (-6+ 7i) ∗12i = -6∗12i + 7i ∗12i

                                   = -72i + 84i²

                                   = -72i + 84(-1)    since i² = -1

                                   = - 84- 72i

Try these questions

Multiply the following complex numbers

1. (3-2i)∗ (3+ 2i)
   Answer: Let z₁ = 3-2i,        a₁ = 3,      b₁ = -2
         z₂ = 3+ 2i,      a₂ = 3,     b₂ = 2
   Comparing with z₁∗ z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)
   (3-2i)∗ (3+ 2i) = [3∗3- (-2)∗2] +i [3∗2 + 3(-2)]
                      = [9 + 4] + i [6-6]
                      = 13 + 0i
                      = 13
   Alternately (3-2i) ∗ (3+2i) = (3)² - (2i)²
                                       = 9 - 4i²
                                       = 9 - 4(-1)    since i²= -1
                                       = 9 + 4
                                       = 13


2. (-7 + 4i) ∗ (-7 - 4i)
   Answer: Let z₁ = -7 + 4i,         a₁= -7,       b₁ = 4
         z₂ = -7 - 4i,         a₂ = -7,       b₂= -4
   Comparing with z₁∗ z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)
          (-7 + 4i) ∗(-7 - 4i) = [(-7) ∗(-7) - (4) ∗(-4)+i [(-7) ∗(-4) + (-7) ∗4]
                                   = (49+ 16) + i (28- 28)
                                  = 65 + 0i
                                 = 65 
   Alternately (-7+ 4i) ∗(-7-4i) = (-7)² -(4i)²
                                          = 49 - 16i²
                                          = 49 - 16 (-1)   since i² = -1
                                          = 49 + 16
                                          = 65


3. (16 - 8i) ∗ (-4i + 7)
   Answer: Let z₁=16 - 8i,      a₁=16,      b₁ = -8
        z₂ = -4i + 7,     a₂=7,        b₂= -4
   Comparing with z₁∗ z₂ = (a₁ a₂ - b₁ b₂ ) +i (a₁ b₂ + a₂ b₁)
        (16 - 8i) ∗ (-4i + 7) = [16 ∗7- (-8) ∗(-4)] +i [16 ∗(-4) + 7 ∗(-8)]
                                  = (112 - 32)+ i (-64 - 56)
                                  = 80 - 120i


4. (-1 + 15i) ∗(15i- 14)
   Answer: Let z₁= -1 + 15i,        a₁= -1,       b₁ =15
        z₂ = -15i- 14,       a₂ = -14,     b₂ = -15
   Comparing with z1∗ z₂ = (a₁ a₂ - b₁ b₂ ) +i (a₁ b₂ + a₂ b₁)
                                        (-1 + 15i) ∗(15i- 14)
                                    = [(-1) ∗(-14) - 15 ∗(-15)] + i [(-1) ∗(-15) + (-14) ∗(15)]
                                    = (14+ 225) + i(15 - 210)
                                    = 239- 195i


5. (-2/3i - 3/4) ∗( 4-3i)
   Answer:  Let z₁=-(2i/3) - 3/4,  a₁=- 3/4,      b₁ = -2/3
        z₂ = 4-3i,           a₂ = 4,          b₂= -3
   Comparing with z₁∗ z₂ = (a₁ a₂- b₁ b₂) +i (a₁ b₂+ a₂ b₁)
                                    = [-(2i/3)-3/4] ∗(4-3i)
                                   = [(-3/4) ∗4 - (-2/3) ∗(-3)] +i [(-3/4) ∗(-3) +4 ∗(-2/3) ]
                                   = (-3-2) +i (+9/4- 8/3)
                                   = -5 +i ((27-32)/12)
                                   = -5 -5/12i