HOW TO MULTIPLY AND DIVIDE COMPLEX NUMBERS

How to Multiply and Divide Complex Numbers ?

Multiplying complex numbers :

Suppose a, b, c, and d are real numbers. Then,

  • (a + bi)(c + di) = (ac − bd) + (ad + bc)i

Division of complex numbers :

To divide the complex number which is in the form

(a + ib)/(c + id)

we have to multiply both numerator and denominator by  the conjugate of the denominator.

That is,

[ (a + ib)/(c + id) ] ⋅ [ (c - id) / (c - id) ]

 =  [ (a + ib) (c - id) / (c + id) (c - id) ]

Example 1 :

Multiply the following complex numbers

(2 + 3i) (4 - 7i)

Solution :

(2 + 3i) (4 - 7i)  =  2(4) + 2(-7i) + 4(3i) + 3i(-7i)

  =  8 - 14i + 12i - 21i2

  =  8 - 2i - 21(-1)

  =  8 - 2i + 21

  =  29 - 2i

Example 2 :

Multiply the following complex numbers

(4 - 2i) (3 - 5i)

Solution :

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

  =  12 - 20i - 6i + 10i2

  =  12 - 26i + 10(-1)

  =  12 - 10 - 26i

  =  2 - 26i

Example 3 :

Multiply the following complex numbers

(-5 + 3i)(-2 + i)

Solution :

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

=  10 - 5i - 6i + 3i2

=  10 - 11i + 3(-1)

=  10 - 3 - 11i

  =  7 - 11i

Example 4 :

Multiply the following complex numbers

(3 - i) (8 + 7i)

Solution :

(3 - i) (8 + 7i)  =  3(8) + 3(7i) - i(8) - i(7i)

=  24 + 21i - 8i - 7i2

=  24 + 13i - 7(-1)

=  24 + 13i + 7

=  31 + 13i 

Example 5 :

Divide the complex number (3 + 2i) by (2 + 4i)

Solution :

(3 + 2i) by (2 + 4i)  =  (3 + 2i)/(2 + 4i)

Whenever we have complex numbers in the denominator, we have to multiply the numerator and denominator by the conjugate of the denominator of the given complex number.

  =  [(3 + 2i)/(2 + 4i)] ⋅[(2 - 4i)/(2 - 4i)]

  =  [(3 + 2i)(2 - 4i)/(2 + 4i) (2 - 4i)]

Multiplying the numerator, we get

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

  =  6 - 3i + 4i - 8i2

  =  6 - 8(-1) + i

  =  6 + 8 + i

  =  14 + i

Multiplying the denominator, we get

(2 + 4i) (2 - 4i)  =  2(2) + 2(-4i) + 4i(2) + 4i(-4i)

  =  4 - 8i + 8i - 16i2

  =  4 - 16(-1)

  =  4 + 16

  =  20

 (3 + 2i)/(2 + 4i)  =  (14 + i)/20

Example 6 :

Which of the following is equal to 

√-1 - √-4 + √-9 ?

a)  i       b)  2i      c)  3i      d) 4i

Solution :

= √-1 - √-4 + √-9

= i - i √4 + i√9

= i - i √(2 x 2) + i√(3 x 3)

= i - 2i + 3i

= 4i - 2i 

= 2i

So, the answer is option b.

Example 7 :

If (4 + i)2 = a + ib, what is the value of a + b ?

Solution :

(4 + i)2 = a + ib

Using algebraic identity (a + b)2 = a2 + 2ab + b2 we get

42 + 2(4)(i) + i2 = a + ib

16 + 8i - 1 = a + ib

15 + 8i = a + ib

Comparing the corresponding terms, we get

a = 15 and b = 8

Example 8 :

If the expression (3 - i) / (1 - 2i) is rewritten in the form a + ib, in which a and b are real numbers, what is the value of a + b?

Solution :

= (3 - i) / (1 - 2i)

Since we have to two complex numbers in the numerator and denominator, we have to multiply both numerator and denominator by the conjugate of the denominator.

Conjugate of 1 - 2i is 1 + 2i

= [(3 - i) / (1 - 2i)] [(1 + 2i) / (1 + 2i)]

= (3 - i)(1 + 2i) / (1 - 2i) (1 + 2i)

= (3 + 6i - i - 2i2) / (12 - (2i)2)

= (3 + 5i - 2(-1)) / (12 - 4i2)

= (3 + 5i + 2) / (1 - 4(-1))

= (5 + 5i) / (1 + 4)

= (5 + 5i) / 5

= (5/5) + (5i/5)

= 1 + i

a = 1, b = 1

a + b = 1 + 1

= 2

Example 9 :

Which of the following complex numbers is equivalent to

(1 - i)2/ (1 + i)

a)  -i/2 - 1/2    b)  -i/2 + 1/2      c)  -i - 1      d)  -i + 1

Solution :

= (1 - i)2/ (1 + i)

= (12 - 2i + i2) / (1 + i)

= (1 - 2i - 1) / (1 + i)

= - 2i / (1 + i)

= [- 2i/(1 + i)] x [(1 - i) / (1 - i)]

= -2i (1 - i) / (1 + i)(1 - i)

= -2i(1 - i) / (12 - i2)

= -2i(1 - i) / (1 + 1)

= -2i(1 - i) / 2

= -i(1 - i)

= -i + i2

= -i - 1

= -1 - i

So, option c is correct.

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