OPERATIONS WITH COMPLEX NUMBERS PRACTICE

Example 1 :

Add the following complex numbers

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

Solution :

To add two complex numbers which is in the form a + ib and c + id, we use the following method.

a + ib + c + id  =  (a + c) + i(b + d)

That is, we have to combine the real part and imaginary part separately.

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

  =  5 - i

Example 2 :

Add the following complex numbers

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

Solution :

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

  =  2 - 2i

Example 3 :

Add the following complex numbers

(-5 + 8i) + (9 - 11i)

Solution :

(-5 + 8i) + (9 - 11i)  =  (-5 + 9) + i(8 - 11)

  =  4 - 3i 

Example 4 :

Add the following complex numbers

(3 + 2i) and (-6 - 9i)

Solution :

(3 + 2i) + (-6 - 9i)  =  (3 - 6) + i(2 - 9)

  =  -3 - 7i

Example 5  :

Subtract

9 - 11i from 2 + 3i

Solution :

  =  (2 + 3i) - (9 - 11i)

  =  (2 - 9) + i(3 - 11)

  =  -7 - 8i

Example 6 :

Subtract

3 + 4i from 4 - 5i

Solution :

  =  (4 - 5i) - (3 + 4i)

  =  4 - 5i - 3 - 4i

  =  (4 - 3) - 5i - 4i

  =  1 - 9i

Example 7 :

Subtract

(-7 + 5i) from (-8 + 9i)

Solution :

  =  -8 + 9i - (-7 + 5i)

  =  -8 + 9i + 7 - 5i

  =  -8 + 7 + 9i - 5i

  =  -1 + 4i

Example 8 :

Subtract

(-11 - 13i) from (-8 - 9i)

Solution :

(-8 - 9i) - (-11 - 13i)  =  -8 - 9i + 11 + 13i

  =  (-8 + 11) + i(-9 + 13)

  =  3 + 4i

Complete any two and write in standard form.

Example 9 :

-6 - (-2 - 2i) - (5 - 4i)

Solution :

= -6 - (-2 - 2i) - (5 - 4i)

Using distributive property, distributing negatives we get

= -6 + 2 + 2i - 5 + 4i

= -6 + 2 - 5 + 2i + 4i

= -9 + 6i

Example 10 :

-4(-7 + 8i)(-5 + 6i)

Solution :

= -4(-7 + 8i)(-5 + 6i)

Using distributive property, distributing negatives we get

= -4(35 - 42i - 40i + 48i²)

= -4(35 - 82i + 48(-1))

= -4(35 - 82i - 48)

= -4(- 82i - 13)

= 328i + 52

Example 11 :

(-3 + i)²

Solution :

= (-3 + i)²

Looks like the algebraic identity,

(a + b)² = a² + 2ab + b²

= (-3)² + 2(-3) i + i²

= 9 - 6i - 1

= 8 - 6i

Example 12 :

(9 + 4i)/6i

Solution :

= (9 + 4i)/6i

Multiplying both numerator and denominator by the conjugate of the denominator, we get

= [(9 + 4i)/6i][-6i/-6i]

= (9 + 4i)(-6i) / -36i²

= (-54i - 24i²) / -36(-1)

= (-54i - 24(-1)) / 36

= (-54i + 24) / 36

= -3i/2 + 2/3

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

Example 13 :

(-5 + 5i) - (4 - 2i) + (-8 - 7i)²

Solution :

= (-5 + 5i) - (4 - 2i) + (-8 - 7i)²

(a - b)² = a² - 2ab + b²

= -5 + 5i - 4 + 2i + (-8)² - 2(-8)(7i) + (7i)²

= -9 + 7i + 64 + 112i + 49i²

= 55 + 119i + 49(-1)

= 55 + 119i - 49

= 6 + 119i

Example 14 :

(4 + i)/(8 - 7i)

Solution :

= (4 + i)/(8 - 7i)

Multiplying both numerator and denominator by the conjugate of the denominator, we get

= [(4 + i)/(8 - 7i)][(8 + 7i)/(8 + 7i)]

= (4 + i)(8 + 7i)/(8 - 7i)(8 + 7i)

= (32 + 28i + 8i + 7i²) / (8² - (7i)²)

= (32 + 36i - 7) / (64 + 49)

= (25 + 36i) / 113

= (25/113) + (36i/113)

Example 15 :

(-7i)(-2 + 7i)(2 + 6i)

Solution :

= (-7i)(-2 + 7i)(2 + 6i)

= (-7i)(-4 - 12i + 14i + 42i²)

= (-7i)(-4 + 2i - 42)

= (-7i)(-46 + 2i)

= 322i - 14i²

= 322i - 14(-1)

= 322i + 14

= 14 + 322i

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