WRITE THE COMPLEX NUMBER IN STANDARD FORM

Example 1 :

3(cos 30˚ - i sin 30˚)

Solution :

Given, z  =  3(cos 30˚ - i sin 30˚)

By using the calculator, we get

z  =  3[√3/2 - i(1/2)]

z  =  3√3/2 - 3/2i

So, the standard form is 3√3/2 - 3/2i.

Example 2 :

8(cos 210˚ + i sin 210˚)

Solution :

Given, z  =  8(cos 210˚ + i sin 210˚)

By using the calculator, we get

z  =  8[-√3/2 + i(-1/2)]

z  =  (-8√3/2) - (8/2i)

z  =  -4√3 - 4i

So, the standard form is -4√3 - 4i.

Example 3 :

5[cos (-60˚) + i sin (-60˚)]

Solution :

Given, z  =  5[cos (-60˚) + i sin (-60˚)]

By using the calculator, we get

z  =  5[1/2 + i(-√3/2)]

z  =  (5/2) - (5√3/2)i

So, the standard form is 5/2 - (5/2)√3i.

Example 4 :

5(cos π/4 + i sin π/4)

Solution :

Given, z  =  5(cos π/4 + i sin π/4)

By using the calculator, we get

z  =  5[√2/2 + i(√2/2)]

z  =  (5/2)√2 + (5/2)√2i

So, the standard form is (5/2)√2 + (5/2)√2i.

Example 5 :

√2(cos 7π/6 + i sin 7π/6)

Solution :

Given, z  =  √2(cos 7π/6 + i sin 7π/6)

By using the calculator, we get

z  =  √2[-√3/2 + i(-1/2)]

z  =  -√6/2 - √2/2i

So, the standard form is -√6/2 - √2/2i.

Example 6 :

√7(cos π/12 + i sin π/12)

Solution :

Given, z  =  √7(cos π/12 + i sin π/12)

By using the calculator, we get

z  =  √7[((√6 + √2)/4) + i((√6 - √2)/4)]

z  =  √7[(√6 + √2)/4] + √7[(√6 - √2)/4]i

So, the standard form is 

√7[(√6 + √2)/4] + √7[(√6 - √2)/4]i

Write your answer in rectangular form when rectangular form is given and in polar form when polar form is given.

Example 7 :

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

Solution :

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

Using distributive property,

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

= 20 - 12i + 20i - 12i²

= 20 + 8i - 12(-1)

= 20 + 8i + 12

= 32 + 8i

Example 8 :

(6 - 2i) / (2 + 4i)

Solution :

(6 - 2i) / (2 + 4i)

= [(6 - 2i) / (2 + 4i)] [(2 - 4i)/(2 - 4i)]

= (6 - 2i)(2 - 4i) / (2 + 4i)/ (2 - 4i)

= (12 - 24i - 4i + 8i²) / (2² - (4i)²)

= (12 - 28i - 8) / (4 + 16)

= (4 - 28i) / 20

= 4/20 - (28i/20)

= 1/5 - 7i/5

Example 9 :

(-1 - 6i)³

Solution :

(-1 - 6i)³

Looks like (a - b)³, then a³ - 3a²b + 3ab² - b³

= (-1)³ - 3(-1)²6i + 3(-1)(6i)² - (-6i)³

= -1 + 3(6i) - 3(36i²) - (-216i³)

i² = -1 and i³ = -i

= -1 + 18i - 108(-1) + 216(-i)

= -1 + 18i + 108 - 216i

= 107 - 198i

Example 10 :

[2 (cos7π/6 + i sin 7π/6)]³

Solution :

= [2 (cos7π/6 + i sin 7π/6)]³

Distributing the power, we get

= 2³ (cos 7π/6 + i sin 7π/6)³

= 8 [cos (3 x 7π/6) + i sin (3 x 7π/6)]

= 8 [cos (7π/2) + i sin (7π/2)]

Write the expression as a complex number in standard form. 

Example 11 :

(3 + 4i ) − (7 − 5i ) + 2i(9 + 12i )

Solution :

= (3 + 4i) − (7 − 5i) + 2i(9 + 12i)

= 3 + 4i - 7 + 5i + 18i + 24i²

= -4 + 9i + 18i + 24(-1)

= -4 - 24 + 27i

= -28 + 27i

Example 12 :

3i(2 + 5i) + (6 − 7i) − (9 + i)

Solution :

= 3i(2 + 5i) + (6 − 7i) − (9 + i)

Using distributive property, we get

= 6i + 15i² + 6 - 7i - 9 - i

= 6i - 7i - i + 15(-1) + 6 - 9

= 6i - 8i - 15 + 6 - 9

= -2i - 24 + 6

= -2i - 18

Example 13 :

(3 + 5i)(2 − 7i⁴) 

Solution :

= (3 + 5i)(2 − 7i⁴)

Using distributive property, we get

= 3(2) + 3(−7i⁴) + 5i(2) + 5i (−7i⁴)

i⁴ = (i²)²

= (-1)²

i⁴ = 1

= 3(2) + 3(−7) + 5i(2) + 5i (−7)

= 6 - 21 + 10i - 35i

= -15 - 25i

Example 14 :

(2 + 4i⁵) + (1 − 9i⁶) − ( 3 + i⁷)

Solution :

= (2 + 4i⁵) + (1 − 9i⁶) − ( 3 + i⁷)

= (2 + 4i) + (1 + 9) − (3 - i)

= (2 + 4i) + 10 − (3 - i)

= 2 + 4i + 10 - 3 + i

= 12 - 3 + 4i + i

= 9 + 5i

Related pages

• Find the cube root of a complex number
  
• Find the nth root of a complex number
  
• Find the indicated power of a complex number
  
• Plot the complex number in the complex plane
  

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