HOW TO EVALUATE THE GIVEN COMPLEX NUMBERS

Evaluate the following if z = 5 - 2i and w = -1 + 3i.

Example 1 :

z + w

Solution :

z + w  =  (5 - 2i) + (-1+ 3i)

  =  (5 - 1) + (-2i + 3i)

  =  4 + i

Example 2 :

z − iw

Solution :

z - iw  =  (5 − 2i) - i(-1+ 3i)

  =  5 − 2i + i - 3i² 

=  5 − 2i + i - 3(-1) 

=  5 − 2i + i + 3

=  (5 + 3) + i(-2 + 1)

=  8 - i

Example 3 :

2z + 3w

Solution :

2z + 3w =  2(5 − 2i) + 3(-1+ 3i)

=  10 - 4i - 3 + 9i

=  (10 - 3) + i(9 - 4)

=  7 + 5i

Example 4 :

zw

Solution :

zw =  (5 − 2i) (-1+ 3i)

=  -5 + 15i + 2i - 6i²

=  -5 + 15i + 2i - 6(-1)

=  (-5 + 6) + i(15 + 2)

=  1 + 17i

Example 5 :

z² + 2zw+ w²

Solution :

z² + 2zw+ w²  =  21 - 20i + 2 + 2( 1 + 17i) + 10 - 6i

  =  21 - 20i + 2 + 34i + 10 - 6i

  =  (21 + 2 + 10) + i(-20 + 34 - 6)

  =  33 + 8i

Example 6 :

(z + w)²

(z + w)²   =  (4 + i)²

  =  4² + i² + 2(4)i

   =  16 - 1 + 8i

  =  15 + 8i

Simplify :

Example 7 :

(4 + 2i) + (-3 – 5i)

Solution :

= (4 + 2i) + (-3 – 5i)

Distributing the positive sign, we get

= 4 + 2i - 3 - 5i

= 4 - 3 + 2i - 5i

= 1 - 3i

Example 8 :

(-3 + 4i) – (5 + 2i)

Solution :

= (-3 + 4i) – (5 + 2i)

Distributing the negative sign, we get

= -3 + 4i - 5 - 2i

= -3 - 5 + 4i - 2i

= -8 + 2i

Example 9 :

(-8 – 7i) – (5 – 4i)

Solution :

= (-8 – 7i) – (5 – 4i)

Distributing the negative sign, we get

= -8 - 7i - 5 + 4i

= -8 - 5 - 7i + 4i

= -13 - 3i

Example 9 :

(3 – 2i)(5 + 4i)

Solution :

= (3 – 2i)(5 + 4i)

Using distributive property, 

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

= 15 + 12i - 10i - 8i²

= 15 + 2i - 8(-1)

= 15 + 2i + 8

= 15 + 8 + 2i

= 23 + 2i

Example 10 :

(3 – 4i)²

Solution :

= (3 – 4i)²

Using the algebraic identity (a - b)² = a² - 2ab + b²

a = 3 and b = 4i

= 3² - 2(3)(4i) + (4i)²

= 9 - 24i + 4²i²

= 9 - 24i + 16(-1)

= 9 - 24i - 16

= 9 - 16 - 24i

= -7 - 24i

Example 10 :

(3 – 2i)(5 + 4i) – (3 – 4i)²

Solution :

= (3 – 2i)(5 + 4i) – (3 – 4i)²

Using distributive property and expanding using algebraic identity, we get

= 3(5) + 3(4i) - 2i(5) - 2i(4i) - [3² - 2(3)(4i) + (4i)²]

= 15 + 12i - 10i - 8i² - [9 - 24i + 16i²]

= 15 + 2i - 8(-1) - [9 - 24i + 16(-1)]

= 15 + 2i + 8 - [9 - 24i - 16]

= 23 + 2i - [-7 - 24i]

= 23 + 2i + 7 + 24i

= 23 + 7 + 2i + 24i

= 30 + 26i

Example 11 :

Multiply √2 + i  into its conjugate.

Solution :

Given complex number is = √2 + i 

Conjugate of complex number = √2 - i 

Multiplying the complex number and its conjugate, we get

=  (√2 + i)(√2 - i)

= √2² - i²

= 2 - (-1)

= 2 + 1

= 3

Example 12 :

Find real x and y , if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i

Solution :

(x – iy) (3 + 5i) is the conjugate of – 6 – 24i

Conjugate of -6-24i is -6 + 24i

(x – iy) (3 + 5i) = -6 + 24i

3x + 5xi - 3yi - 5i² = -6 + 24i

3x + i(5x - 3y) - 5(-1) = -6 + 24i

3x + 5 + i(5x - 3y) = -6 + 24i

Equating the real and imaginary parts, we get

3x + 5 = -6

3x = -6 - 5

3x = - 11

x = -11/3

5x - 3y = 24

Applying the value of x, we get

5(11/3) - 3y = 24

55/3 - 3y = 24

(55/3) - 24 = 3y

3y = (55 - 72)/3

= -17/3

y = -17/9

So, the values of x and y are -11/3 and -17/9 respectively.

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