HOW TO ADD SUBTRACT MULTIPLY AND DIVIDE COMPLEX NUMBERS

Let z₁ and z₂ be two complex numbers such that

z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂,

where x₁, x₂, y₁ and y₂ are real values.

In each of the following examples, write the expression in the form a + bi, where a and b are real numbers.

Example 1 :

(4 + 2i) + (3 + 8i)

Solution :

= (4 + 2i) + (3 + 8i)

= (4 + 3) + (2i + 8i)

= 7 + 10i

Example 2 :

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

Solution :

= (5 + 7i) + (4 + 6i)

= (5 + 4) + (7i + 6i)

= 9 + 15i

Example 3 :

(5 + 3i) - (2 + 9i)

Solution :

= (5 + 3i) - (2 + 9i)

= 5 + 3i - 2 - 9i

= 3 - 6i

Example 4 :

(9 + 2i) - (6 + 7i)

Solution :

= (9 + 2i) - (6 + 7i)

= 9 + 2i - 6 - 7i

= 3 - 5i

Example 5 :

(9 + 2i) - (6 + 7i)

Solution :

= (9 + 2i) - (6 + 7i)

= 9 + 2i - 6 - 7i

= 3 - 5i

Example 6 :

(8 - 4i)(2 - 3i)

Solution :

= (8 - 4i)(2 - 3i)

= 16 - 24i - 8i + 12i²

= 16 - 32i + 12(-1)

= 16 - 32i - 12

= 4 - 32i

Example 7 :

(2 + 3i)(1 - 4i)

Solution :

= (2 + 3i)(1 - 4i)

= 2 - 8i + 3i - 12i²

= 2 - 5i - 12(-1)

= 2 - 5i + 12

= 14 - 5i

Example 8 :

Solution :

Example 9 :

Solution :

Example 10 :

Find the values of the real numbers x and y, if the following two complex numbers are equal.

(3 - i)x - (2 - i)y + 2i + 5 and 2x + (-1 + 2i)y + 3 + 2i

Solution :

(3 - i)x - (2 - i)y + 2i + 5 = 2x + (-1 + 2i)y + 3 + 2i

3x - ix - 2y + iy + 2i + 5 = 2x - y + 2yi + 3 + 2i

3x - 2y + 5 - ix + iy + 2i = 2x - y + 3 + 2yi + 2i

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

The complex numbers sides are in standard form a + ib on both sides.

Equating the real parts and imaginary parts,

(1) + (2) :

-2y + 2 = 0

-2y = -2

y = 1

Substitute y = 1 in (2).

-x - 1 = 0

-x = 1

x = -1

Therefore,

x = -1 and y = 1

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