Complex Numbers Multiplication

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