Maths sheet 1 (PDF)




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Title: Complex Numbers sheet
Author: mufic

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Complex Numbers sheet
1) Find the output of the following operations performed on complex numbers
2+๐‘–
๐‘Ž)
๐‘–+1
d (

2

โˆš3โˆ’ ๐‘–

g)

10

+

1
๐‘–

10

)

(2 + ๐‘–)(1 โˆ’ ๐‘–)
3โˆ’2๐‘–

j) (โˆ’2 โˆ’ 3 ๐‘–) (โˆ’3 + 4๐‘–)

(1โˆ’๐‘– )6

b) (1 โˆ’ โˆš3 ๐‘–)

c)

e) (1 + ๐‘–)5

f) (1 + โˆš3 ๐‘–) (๐‘– โˆ’ 1)7

h)

1
๐‘–(3+2๐‘–)2
(โˆ’2โˆ’3 ๐‘–)

k)

(โˆ’3+4๐‘–)

(๐‘–+1)7
5

i)

l)

(โˆš3+โˆš2 ๐‘–)3
(โˆš2โˆ’โˆš3๐‘–)
1
(โˆ’3+4๐‘–)

2) Find the modulus ( |Z |) and the argument ( arg(Z) ) for next complex numbers

3)

๐‘Ž) ๐‘ = (1 โˆ’ ๐‘–)

b) ๐‘ = 11๐‘–

c) ๐‘ = โˆ’

d) ๐‘ = (โˆ’2 + 2โˆš3๐‘–)

e) ๐‘ = โˆ’4๐‘–

f) ๐‘ =

๐œ‹
4

1
2

Find the roots of complex numbers where

๐‘Ž) ๐‘ 3 = โˆ’8

d)

๐‘ 4 = (โˆ’2 โˆ’ 2โˆš3๐‘–)

b) ๐‘ 3 = 27๐‘–

e)

๐‘ 3 = 4โˆš2๐‘– โˆ’ 4โˆš2

1

c) ๐‘ 2 = (โˆ’1 + โˆš3 ๐‘–)

f) ๐‘ 6 = 64๐‘–

Complex Numbers sheet
g) ๐‘ 6 = โˆ’64๐‘–
j) ๐‘ 3 =

โˆš2
2

โˆ’

4

h) ๐‘ 8 = โˆ’16

โˆš2
2

1
2

i) ๐‘ = โˆ’

โˆš3

๐‘–

4) Given Z= cos(3) + sin(3) ๐‘– prove that 1 + ๐‘ง = (1 + ๐‘ง)๐‘ง
5) Calculate (cos(2) + sin(2)๐‘– + 1)๐‘›
6) Given : n is a positive integer
Z is a complex number with modulus 1, such that ๐‘ง 2๐‘› โ‰  โˆ’1
show that

๐‘ง๐‘›

1+๐‘ง 2๐‘›

is a real number.

7) Let ๐‘ง the conjugate complex number of z. find z such that

๐‘ง 2 + (๐‘ง)2 = ๐‘ง๐‘’๐‘Ÿ๐‘œ
8) Find the value of k for the quotient

(2โˆ’๐‘˜๐‘–)
(๐‘˜โˆ’๐‘–)

if it is :

- A pure imaginary number
- A real number
9) The complex number, 2 + 2
is rotated 45ยฐ about the origin of its
coordinates in an anti-clockwise direction. Find the complex number
obtained after the turn.

10)

Determine the value of b for the quotient

, if it equals:

Note the following trigonometric Identities
a) cos 2 ๐‘ฅ + sin2 ๐‘ฅ = 1
b) cos(2๐‘ฅ) = cos 2 ๐‘ฅ โˆ’ sin2 ๐‘ฅ = 2 cos 2 ๐‘ฅ โˆ’ 1 = 1 โˆ’ 2 sin2 ๐‘ฅ
c) sin(2๐‘ฅ) = 2sin(๐‘ฅ)cos(๐‘ฅ)

2

2

๐‘–






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