Download Presentation of complex number in Cartesian and polar coordinate

1
∎ 𝑎 + 𝑏𝑖 ,
𝑎𝜖𝑅,
∎ 𝑧 = 𝑎 + 𝑏𝑖
𝑏𝜖𝑅
𝑖 = √−1
𝑎 = 𝑅𝑒(𝑧),
∎ 𝑔𝑖𝑣𝑒𝑛 𝑧 = 𝑎 + 𝑏𝑖
𝑏 = 𝐼𝑚(𝑧)
& 𝑤 = 𝑐 + 𝑑𝑖,
𝑎 + 𝑏𝑖 = 𝑐 + 𝑑𝑖
⟺ 𝑎=𝑐 & 𝑏=𝑑
∎ 𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑜𝑓 𝑧 = 𝑎 + 𝑏𝑖 𝑖𝑠 𝑧 ∗ = 𝑎 − 𝑏𝑖
∎ 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒𝑠 𝑖𝑠 𝒓𝒆𝒂𝒍: 𝑧 + 𝑧 ∗ = 2𝑎
∎ 𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒𝑠 𝑖𝑠 𝒓𝒆𝒂𝒍: 𝑧 𝑧 ∗ = 𝑎2 + 𝑏 2
Properties of complex conjugates:
1
1
● 𝑅𝑒(𝑧) = (𝑧 + 𝑧 ∗ )
𝐼𝑚(𝑧) = (𝑧 − 𝑧 ∗ )
2
2
● (𝑧1 ± 𝑧2 )∗ = 𝑧1∗ ± 𝑧2∗
● (𝑧 ∗ )∗ = 𝑧
● (𝑧1 ± 𝑧2 )∗ = 𝑧1∗ ± 𝑧2∗
𝑧1 ∗ 𝑧1∗
● (𝑧1 𝑧2 )∗ = 𝑧1∗ × 𝑧2∗
& ( ) = ∗ , 𝑧2 ≠ 0
𝑧2
𝑧2
● (𝑧 𝑛 )∗ = (𝑧 ∗ )𝑛
Presentation of complex number in Cartesian and polar coordinate system
▪ 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒: |𝑧| = 𝑟 = √𝑥 2 + 𝑦 2
▪ 𝐴𝑟𝑔𝑢𝑚𝑒𝑛𝑡: 𝑎𝑟𝑔 𝑧 = 𝜃 = 𝑎𝑟𝑐 𝑡𝑎𝑛
𝑦
𝑥
𝑏𝑒 𝑐𝑎𝑟𝑒𝑓𝑢𝑙
▪ 𝑥 = 𝑟𝑐𝑜𝑠 𝜃, 𝑦 = 𝑟𝑠𝑖𝑛 𝜃
𝒛 = 𝒙⏟+ 𝒚𝒊 = ⏟
𝒓(𝒄𝒐𝒔 𝜽 + 𝒊 𝒔𝒊𝒏 𝜽)
𝑪𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏
𝒇𝒐𝒓𝒎
𝒑𝒐𝒍𝒂𝒓 𝒇𝒐𝒓𝒎
𝒎𝒐𝒅𝒖𝒍𝒖𝒔−𝒂𝒓𝒈𝒖𝒎𝒆𝒏𝒕
𝒇𝒐𝒓𝒎
Argan plane is the complex plane
𝒓⏟
𝑒𝑖𝜃
=
= 𝒓 𝑐𝑖𝑠 𝜃
Recall that the argument should
be measured in radians
𝑬𝒖𝒍𝒆𝒓 𝒇𝒐𝒓𝒎
Very useful for fast conversions from
Cartesian into Euler form
it will give you visually position of the point,
and therefore quadrant for the angle
𝑧 = –𝑖
3𝜋
– 2 𝑖 = 2𝑒 𝑖 2
1 = 𝑒 𝑖0
→
3𝜋
𝑧 = 𝑒𝑖 2
𝜋
𝑖 = 𝑒𝑖2
2
Practicality of Euler’s form
𝑧1 = 𝑥1 + 𝑖𝑦1 = 𝑟1 (cos 𝜃1 + 𝑖 sin 𝜃1 )
&
𝑧2 = 𝑥2 + 𝑖𝑦2 = 𝑟2 (cos 𝜃2 + 𝑖 sin 𝜃2 )
Product is:
𝑧1 𝑧2 = (𝑥1 + 𝑖𝑦1 )(𝑥2 + 𝑖𝑦2 ) = 𝑟1 𝑟2 (cos 𝜃1 + 𝑖 sin 𝜃1 )(cos 𝜃2 + 𝑖 sin 𝜃2 )
= 𝑟1 𝑟2 (cos 𝜃1 cos 𝜃2 − sin 𝜃1 sin 𝜃2 + 𝑖 (cos 𝜃1 sin 𝜃2 + sin 𝜃1 cos 𝜃2 )
= 𝑟1 𝑟2 [cos(𝜃1 + 𝜃2 ) + 𝑖 sin(𝜃1 + 𝜃2 )]
|𝒛𝟏 𝒛𝟐 | = |𝒛𝟏 ||𝒛𝟐 |
∴
𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒊𝒔 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒎𝒐𝒅𝒖𝒍𝒖𝒔
& 𝒂𝒓𝒈 (𝒛𝟏 𝒛𝟐 ) = 𝒂𝒓𝒈 (𝒛𝟏 ) + 𝒂𝒓𝒈 (𝒛𝟐 )
𝒂𝒓𝒈𝒎𝒆𝒏𝒕 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒊𝒔 𝒔𝒖𝒎 𝒐𝒇 𝒂𝒓𝒈𝒖𝒎𝒆𝒏𝒕𝒔
▪ 𝒛𝟏 𝒛𝟐 = [|𝒛𝟏 |𝒆𝒊𝜽𝟏 ] [|𝒛𝟐 |𝒆𝒊𝜽𝟐 ] = |𝒛𝟏 ||𝒛𝟐 |𝒆𝒊(𝜽𝟏+𝜽𝟐)
Quotient is:
(𝑥1 + 𝑖𝑦1 ) 𝑟1 cos 𝜃1 + 𝑖 sin 𝜃1 𝑟1 cos 𝜃1 + 𝑖 sin 𝜃1 cos 𝜃2 − 𝑖 sin 𝜃2
𝑧1
=
=
=
×
(𝑥
𝑧2
𝑟2 cos 𝜃2 + 𝑖 sin 𝜃2 𝑟2 cos 𝜃2 + 𝑖 sin 𝜃2 cos 𝜃2 − 𝑖 sin 𝜃2
2 + 𝑖𝑦2 )
=
∴
▪
𝑟1 cos 𝜃1 cos 𝜃2 + sin 𝜃1 sin 𝜃2 + 𝑖 (sin 𝜃1 𝑐𝑜𝑠 𝜃2 − cos 𝜃1 sin 𝜃2 )
𝑟1
= [cos( 𝜃1 − 𝜃2 ) + 𝑖 sin( 𝜃1 − 𝜃2 )]
2
2
𝑟2
cos 𝜃2 + sin 𝜃2
𝑟2
|𝒛𝟏 |
𝒛𝟏
| |=
|𝒛𝟐 |
𝒛𝟐
𝒛𝟏
& 𝒂𝒓𝒈 ( ) = 𝒂𝒓𝒈 (𝒛𝟏 ) − 𝒂𝒓𝒈 (𝒛𝟐 )
𝒛𝟐
𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒊𝒔 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒐𝒇 𝒎𝒐𝒅𝒖𝒍𝒖𝒔
𝒂𝒓𝒈𝒎𝒆𝒏𝒕 𝒐𝒇 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒊𝒔 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒂𝒓𝒈𝒖𝒎𝒆𝒏𝒕𝒔
𝒛𝟏
[|𝒛𝟏 |𝒆𝒊𝜽𝟏 ] | 𝒛𝟏 | 𝒊(𝜽 −𝜽 )
=
=
𝒆 𝟏 𝟐
𝒛𝟐
[|𝒛𝟐 |𝒆𝒊𝜽𝟐 ] | 𝒛𝟐 |
Conclusion: Euler form of complex numbers follows ordinary algebra: 𝑎𝑛 𝑎𝑚 = 𝑎𝑛+𝑚
25
𝐼𝑡 𝑖𝑠 𝑒𝑎𝑠𝑦 𝑡𝑜 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑤𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑎𝑛𝑦 𝑓𝑜𝑟𝑚, 𝑏𝑢𝑡 𝑤ℎ𝑎𝑡 𝑖𝑓 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 10 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑟 𝑧 25 𝑜𝑟 √𝑧
Properties of modulus and argument
▪ |𝑧 ∗ | = |𝑧|
& 𝑎𝑟𝑔 (𝑧 ∗ ) = −𝑎𝑟𝑔 𝑧
▪ 𝑧𝑧 ∗ = |𝑧|2
▪ 𝑒 𝑖𝜃 = 𝑒 𝑖(𝜃+𝑘2𝜋)
𝑘𝜖𝑍
3
De Moivre’s Theorem
𝑛
𝑧 𝑛 = [𝑟(𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)]𝑛 = 𝑟 𝑛 (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)𝑛 = [𝑟𝑒 𝑖𝜃 ] = 𝑟 𝑛 𝑒 𝑖𝑛𝜃 = 𝑟 𝑛 (𝑐𝑜𝑠 𝑛𝜃 + 𝑖 𝑠𝑖𝑛 𝑛𝜃)
𝑛
(|𝑧|𝑒 𝑖𝜃 ) = |𝑧|𝑛 𝑒 𝑖𝑛𝜃
∴ (𝒄𝒐𝒔 𝒏𝜽 + 𝒊 𝒔𝒊𝒏 𝒏𝜽) = (𝒄𝒐𝒔𝜽 + 𝒊 𝒔𝒊𝒏𝜽)𝒏
Application of DeMoivre’s Theorem
(𝑐𝑜𝑠 𝑛𝜃 + 𝑖 𝑠𝑖𝑛 𝑛𝜃) = (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)𝑛
𝑛
𝑛
𝑛
𝑛
(𝑎 + 𝑏)𝑛 = ∑ ( ) 𝑎𝑛−𝑟 𝑏 𝑟 = 𝑎𝑛 + ( ) 𝑎𝑛−1 𝑏 + ⋯ + ( ) 𝑎𝑛−𝑟 𝑏 𝑟 + ⋯ + 𝑏 𝑛
𝑟
1
𝑟
𝑟=0
𝑛(𝑛 − 1)(𝑛 − 2) ⋯ (𝑛 − 𝑟 + 1)
𝑛
( )=
𝑟
𝑟!
=
𝑛!
𝑛!
𝑛
=
=(
)
𝑛−𝑟
(𝑛 − 𝑟)! 𝑟!
𝑟! (𝑛 − 𝑟)!
𝑛
( ) ≡1
0
Certain trigonometric identities can be derived using DeMoivre’s theorem. We can for instance express cos n ,
sin n and tan n in terms of cos  , sin  and tan  .
Example:
We can find an expression for cos5  Re(cos5  i sin 5 )
= Re  cos   i sin  
5
Then:
(using DeMoivre’s theorem)
cos 5𝜃 = 𝑅𝑒 (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)5
= 𝑅𝑒(𝑐𝑜𝑠 5 𝜃 + 5𝑖 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 + 10𝑖 2 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛2 𝜃 + 10𝑖 3 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 5𝑖 4 cos 𝜃 𝑠𝑖𝑛4 𝜃 + 𝑖 5 𝑠𝑖𝑛5 𝜃)
= 𝑅𝑒(𝑐𝑜𝑠 5 𝜃 + 5𝑖 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 + 10𝑖 2 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛2 𝜃 + 10𝑖 3 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 5𝑖 4 cos 𝜃 𝑠𝑖𝑛4 𝜃 + 𝑖 5 𝑠𝑖𝑛5 𝜃)
= 𝑐𝑜𝑠 5 𝜃 − 10 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛2 𝜃 + 5 cos 𝜃 𝑠𝑖𝑛4 𝜃
cos5  cos5   10cos3  sin 2   5cos  sin 4  .
(*)
If required, the right hand side can be expressed entirely in terms of cos  . We get:
cos5  cos5   10cos3  (1  cos2  )  5cos  (1  cos 2  )2
= cos5   10cos3   10cos5   5cos  10cos3   5cos5 
cos5  16cos5   20cos3   5cos 
4
Note 1: We can also get an identity for sin 5 :
sin 5𝜃 = 𝐼𝑚 (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)5
= 𝐼𝑚(𝑐𝑜𝑠 5 𝜃 + 5𝑖 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 + 10𝑖 2 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛2 𝜃 + 10𝑖 3 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 5𝑖 4 cos 𝜃 𝑠𝑖𝑛4 𝜃 + 𝑖 5 𝑠𝑖𝑛5 𝜃)
= 𝐼𝑚(𝑐𝑜𝑠 5 𝜃 + 5𝑖 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 + 10𝑖 2 𝑐𝑜𝑠 3 𝜃 𝑠𝑖𝑛2 𝜃 + 10𝑖 3 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 5𝑖 4 cos 𝜃 𝑠𝑖𝑛4 𝜃 + 𝑖 5 𝑠𝑖𝑛5 𝜃)
= 5 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 − 10 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 𝑠𝑖𝑛5 𝜃
sin 5𝜃 = 5 𝑐𝑜𝑠 4 𝜃 𝑠𝑖𝑛𝜃 − 10 𝑐𝑜𝑠 2 𝜃 𝑠𝑖𝑛3 𝜃 + 𝑠𝑖𝑛5 𝜃
(**)
If required, the right hand side can be expressed entirely in terms of sin 
sin 5  5(1  sin 2  )2 sin   10(1  sin 2  )sin 3   sin 5 
= 5sin   10sin3   5sin5   10sin3   10sin5   sin5 
sin 5  16sin 5   20sin 3   5sin 
Note 2: We can also get an expression for tan 5 by dividing equation (*) by equation (**):
sin 5 5cos 4  sin   10 cos 2  sin 3   sin 5 
tan 5 

cos 5 cos5   10 cos3  sin 2   5cos  sin 4 
Dividing every term on the top and bottom by cos5  gives:
tan 5

5cos 4  sin  10 cos 2  sin 3  sin 5 


cos5 
cos5 
cos5 
cos5  10 cos3  sin 2  5cos  sin 4 


cos5 
cos5 
cos5 
=
tan   10 tan 3   tan 5 
1  10 tan 2   5 tan 4 
5
Question:
a) Find an expression for cos 4 in terms of cos  only.
b) Find an expression for sin 4 in terms of sin  only.
c) Show that tan 4 
4t  4t 3
1  6t 2  t 4
, where t = tanθ.
Questions:
1. Find an expression for cos 6 in terms of c  cos and s  sin 
2. Find an expression for sin 7 in terms of sin  only.
3. Find an expression for tan 7 in terms of t  tan  .
Finding a general root of a complex number
n
General problem: Find the complex numbers z such that z =a + ib.
Example: Find the cube roots of 8 – 8i, i.e. find z such that z3 = 8 – 8i.
The nth roots of the complex number c are n solutions of zn = c. There are exactly n nth roots of c.
𝑙𝑒𝑡 𝑐 = |𝒄|𝑒 𝑖𝜃 = |𝒄|𝑒 𝑖(𝜃+𝑘2𝜋)
𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑧 𝑛 = 𝑐 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝑧 = √|𝑐| 𝑒 𝑖
𝑛
(𝜃+𝑘2𝜋)
𝑛
𝑛
= √|𝑐| {𝑐𝑜𝑠 (
𝜃+𝑘2𝜋
)+𝑖
𝑛
𝜃+𝑘2𝜋
)}
𝑛
𝑠𝑖𝑛 (
𝑘 = 0, 1, 2, 3 … , 𝑛 − 1
Only the values k = 0, 1, …, n - 1 give different values of z
Geometrically, the n th roots are the vertices
of a regular polygon with n sides in Argan plane.
zn = 1
2
Example: Find 5th root of 1. Or show that if   e 5 i , then the 5th roots of unity can be expressed as
1,  ,  2 ,  3 ,  4
1 = 𝑒 𝑖(0+𝑘2𝜋)
𝑖
√1 = 𝑒
5
(0+𝑘2𝜋)
5
1
𝑘𝜖𝑁
𝑘 = 0, 1, 2, 3, 4
-1
So the 5th roots of unity are 1e
0
, 1e
2
5
i
, 1e
4
5
i
, 1e
6
5
i
, 1e
8
5
1
i
-1
6
Example: Find the cube roots of 8 – 8i, i.e. find z such that z3 = 8 – 8i. 8 – 8i = [ 128 , 4 ]
𝑐 = |𝒄|𝑒 𝑖𝜃 = |𝒄|𝑒 𝑖(𝜃+𝑘2𝜋) so
[ 128 , 4 ] or [ 128 , 74 ] or [ 128 , 158 ]
Let z = [r, θ] be a cube root of 8 – 8i.
Then z 3  [r 3 ,3 ] = [ 128 , 4 ] or [ 128 , 74 ] or [ 128 , 154 ]
Comparing the modulus and arguments we get:

r 3  128 i.e. r  128
3   4 or
7
4
or
15
4
1 / 3  (128)1 / 6  2.244924097
 or 7 or
so    12
12
15
12
The cube roots of 8 – 8i are:
6
 )  i sin(  ))
128(cos( 12
12
6
128(cos( 712 )  i sin( 712 ))
6
 )  i sin( 15 )) The cube roots (to 3 sf) are:
128(cos( 15
12
12
2.17 - 0.581i
-0.581 + 2.17i
-1.59 – 1.59iNote: The cube roots of 8 – 8i can be shown on an Argand diagram:
3
Notice that the cube roots form an equilateral triangle.
2
1
-2
-1
0
-1
-2
1
2
3
7
REAL POLYNOMIALS are polynomials with real coefficients.
REMAINDER
8
9
P(x) is real and – 3 + i is zero. ∴ – 3 – i
is zero
[𝑎𝑥 3 + 9𝑥 2 + 𝑎𝑥 − 30] = [𝑥 − (−3 + 𝑖)][𝑥 − (−3 − 𝑖)](𝑎𝑥 + 𝑏)
[𝑎𝑥 3 + 9𝑥 2 + 𝑎𝑥 − 30] = (𝑥 2 + 6𝑥 + 10)(𝑎𝑥 + 𝑏) = 𝑎𝑥 3 + (6𝑎 + 𝑏)𝑥 2 + (10𝑎 + 6𝑏)𝑥 + 10𝑏
∴ 6𝑎 + 𝑏 = 9
&
10𝑏 = −30
10𝑎 + 6𝑏 = 𝑎
10𝑏 = −30 → 𝑏 = −3
10𝑎 + 6𝑏 = 𝑎 → 9𝑎 + 6𝑏 = 0 → 𝑎 = 2
𝑜𝑟 6𝑎 + 𝑏 = 9 → 𝑎 = 2
linear factor (ax+b) = 2x – 3 so zeroes are:
−3 ± 𝑖
𝑎𝑛𝑑
3
2
Complex Numbers
1.
Let z = x + yi. Find the values of x and y if (1 – i)z = 1 – 3i.
2.
(a)
Evaluate (1 + i)2, where i =
(b)
Prove, by mathematical induction, that (1 + i)4n = (–4)n, where n 
(c)
Hence or otherwise, find (1 + i)32.
[4]
1 .
*.
[10]
10
3.
Let z1 =
6 i 2
, and z2 = 1 – i.
2
(a)
Write z1 and z2 in the form r(cos θ + i sin θ), where r > 0 and –
(b)
Show that
(c)
Find the value of
z1


= cos
+ i sin
.
12
12
z2
z1
in the form a + bi, where a and b are to be determined exactly in radical
z2
(surd) form. Hence or otherwise find the exact values of cos
4.
π
π
θ .
2
2


and sin
.
12
12
[12]
Let z1 = a  cos   i sin   and z2 = b  cos   i sin  .
3
3
4
4


3
z 
Express  1  in the form z = x + yi.
 z2 
[3]
5.
If z is a complex number and |z + 16| = 4 |z + l|, find the value of | z|.
[3]
6.
Find the values of a and b, where a and b are real, given that (a + bi)(2 – i) = 5 – i.
[3]
7.
Given that z = (b + i)2, where b is real and positive, find the exact value of b when arg z = 60°.
[3]
8.
The complex number z satisfies i(z + 2) = 1 – 2z, where i  – 1 . Write z in the form z = a + bi, where a
and b are real numbers.
9.
The complex number z satisfies the equation
2
z=
+ 1 – 4i.
1– i
Express z in the form x + iy where x, y  .
[3]
[5]
10.
Consider the equation 2(p + iq) = q – ip – 2 (1 – i), where p and q are both real numbers. Find p and q.
[6]
11.
Let the complex number z be given by
z=1+
i
.
i– 3
Express z in the form a +bi, giving the exact values of the real constants a, b.
[6]
11
12.
z  z  3i .
A complex number z is such that
(a)
(b)
(c)
3
.
2
Let z1 and z2 be the two possible values of z, such that z  3.
(i)
Sketch a diagram to show the points which represent z1 and z2 in the complex plane, where
z1 is in the first quadrant.
π
(ii) Show that arg z1 = .
6
(iii) Find arg z2.
 zk z 
Given that arg  1 2  = π, find a value of k.
 2i 


Show that the imaginary part of z is
[10]
13.
Given that (a + i)(2 – bi) = 7 – i, find the value of a and of b, where a, b 
.
[6]
14.
Given that z 
, solve the equation z3 – 8i = 0, giving your answers in the form z = r (cos + i sin).
[6]
15.
Given that z = (b + i)2, where b is real and positive, find the exact value of b when arg z = 60°.
[6]
16.
Given that | z | = 2 5 , find the complex number z that satisfies the equation
25  15  1  8i.
z
z*
7.
The two complex numbers z1 =
[6]
b
a
and z2 =
where a, b
1 2i
1i
, are such that z1 + z2 = 3. Calculate
the value of a and of b.
[6]
18.
19.
The complex numbers z1 and z2 are z1 = 2 + i, z2 = 3 + i.
(a)
Find z1z2, giving your answer in the form a + ib, a, b
(b)
The polar form of z1 may be written as  5 ,arctan  .


1
2
(i)
Express the polar form of z2, z1 z2 in a similar way.
(ii)
Hence show that


π
4
Let z1 = r  cos  i sin
1
π
1
= arctan + arctan .
3
4
2
π
 and z2 = 1 +
4
(a)
Write z2 in modulus-argument form.
(b)
Find the value of r if z1 z 2
3
= 2.
.
[6]
3 i.
[6]
12
20.
Let z1 and z2 be complex numbers. Solve the simultaneous equations
2z1 + z2 = 7, z1 + iz2 = 4 + 4i
Give your answers in the form z = a + bi, where a, b
21.
.
[6]
The complex number z is defined by
2π
2π 
π
π


z = 4  cos  i sin   4 3  cos  i sin .
3
3 
6
6


22.
(a)
Express z in the form rei, where r and  have exact values.
(b)
Find the cube roots of z, expressing in the form rei, where r and  have exact values.
The polynomial P(z) = z3 + mz2 + nz −8 is divisible by (z +1+ i), where z
value of m and of n.
[6]
and m, n . Find the
[6]
23.
Let u =1+
3 i and v =1+ i where i2 = −1.
(a)
(i)
Show that
(ii)
By expressing both u and v in modulus-argument form show that
3 1
3 1
u


i.
v
2
2
u
π
π

 2  cos  i sin 
v
12
12 

.
π
in the form a  b 3 where a, b
12
Use mathematical induction to prove that for n +,
(iii)
(b)
Hence find the exact value of tan
1 3 i  2
n
(c)
Let z =
n
.
nπ
nπ 

 cos  i sin .
3
3 

2 vu
2 v u
.
Show that Re z = 0.
24.
(a)
[28]
Express the complex number 1+ i in the form
ae
i
π
b
, where a, b
+
.
n
(b)
 1 i 
 , where n
Using the result from (a), show that 
 2
(c)
Hence solve the equation z8 −1 = 0.
, has only eight distinct values.
25.
Find, in its simplest form, the argument of (sin + i (1− cos ))2 where  is an acute angle.
26.
Consider w =
z
where z = x + iy, y  0 and z2 + 1  0.
z 1
2
[9]
[7]
13
Given that Im w = 0, show that z = 1.
[7]
27.
(z + 2i) is a factor of 2z3–3z2 + 8z – 12. Find the other two factors.
[3]
28.
Let P(z) = z3 + az2 + bz + c, where a, b, and c 
(–3 + 2i). Find the value of a, of b and of c.
[6]
. Two of the roots of P(z) = 0 are –2 and
De Moivre’s Theorem
1.
2.
Let x and y be real numbers, and  be one of the complex solutions of the equation z3 = 1. Evaluate:
(a)
1 +  + 2;
(b)
( x + 2y)(2x +  y).
(a)
Express z5 – 1 as a product of two factors, one of which is linear.
(b)
Find the zeros of z5 – 1, giving your answers in the form
[6]
r(cos θ + i sin θ) where r > 0 and –π < θ  π.
3.
(c)
Express z4 + z3 + z2 + z + 1 as a product of two real quadratic factors.
(a)
Express the complex number 8i in polar form.
(b)
The cube root of 8i which lies in the first quadrant is denoted by z. Express z
(i)
in polar form;
(ii)
in cartesian form.
[6]
2
4.
Consider the complex number z =
(a)
5.
[10]
π
π 
π
π

 cos – i sin   cos  i sin 
4
4 
3
3

π
π 

– i sin 
 cos
24
24 

4
(i)
Find the modulus of z.
(ii)
Find the argument of z, giving your answer in radians.
3
.
(b)
Using De Moivre’s theorem, show that z is a cube root of one, ie z = 3 1 .
(c)
Simplify (l + 2z)(2 + z2), expressing your answer in the form a + bi, where a and b are exact real
numbers.
(a)
Prove, using mathematical induction, that for a positive integer n,
(cos + i sin)n = cos n + i sin n where i2 = –1.
(b)
The complex number z is defined by z = cos + i sin.
[11]
14
(c)
1
= cos (–) + i sin (–).
z
(i)
Show that
(ii)
Deduce that zn + z–n = 2 cos nθ.
(i)
Find the binomial expansion of (z + z–l)5.
(ii)
Hence show that cos5 =
1
(a cos 5 + b cos 3 + c cos ),
16
where a, b, c are positive integers to be found.
6.
(a)
Use mathematical induction to prove De Moivre’s theorem
(cos + i sin)n = cos (n) + i sin (n), n 
(b)
[15]
+
.
Consider z5 – 32 = 0.
(i)

 2π 
 2π  
Show that z1 = 2  cos    i sin    is one of the complex roots of this equation.
 5 
 5 

(ii)
Find z12, z13, z14, z15, giving your answer in the modulus argument form.
(iii)
Plot the points that represent z1, z12, z13, z14 and z15, in the complex plane.
(iv)
The point z1n is mapped to z1n+1 by a composition of two linear transformations, where n =
1, 2, 3, 4. Give a full geometric description of the two transformations.
7.
Given that z 
[16]
, solve the equation z3 – 8i = 0, giving your answers in the form z = r (cos + i sin).
[6]
8.
Consider the complex number z = cos + i sin.
(a)
Using De Moivre’s theorem show that
zn +
1
= 2 cos n.
zn
4
(b)
1

By expanding  z   show that
z

cos4 =
(c)
Let g (a) =

a
0
1
(cos 4 + 4 cos 2 + 3).
8
cos4 d .
(i)
Find g (a).
(ii)
Solve g (a) = 1
[11]
15
9.
Let z = cos  + i sin , for –
(a)
π
π
 .
4
4
(i)
Find z3 using the binomial theorem.
(ii)
Use de Moivre’s theorem to show that
cos 3 = 4 cos3 – 3 cos and sin 3 = 3 sin – 4 sin3.
10.
(b)
Hence prove that
sin 3θ  sin θ
= tan.
cos 3θ  cos θ
(c)
Given that sin =
1
, find the exact value of tan 3.
3
[21]
Let y = cos + i sin.
(a)
dy
= iy.
dθ
Show that
[You may assume that for the purposes of differentiation and integration, i may be treated in the
same way as a real constant.]
(b)
Hence show, using integration, that y = ei.
(c)
Use this result to deduce de Moivre’s theorem.
(d)
(i)
Given that
sin 6θ
= a cos5 + b cos3 + c cos, where sin  0, use de Moivre’s theorem
sin θ
with n = 6 to find the values of the constants a, b and c.
Hence deduce the value of lim
(ii)
 0
sin 6θ
.
sin θ
11.
Prove by induction that 12n + 2(5n−1) is a multiple of 7 for n 
12.
Prove that
13.
Express
14.
Let w = cos
15.
 3  i   3  i
n
1
1 i 3 
3
n
is real, where n
+
+
.
[10]
.
a
where a, b .
b
in the form
[20]
[6]
[5]
2
2
 i sin .
5
5
(a)
Show that w is a root of the equation z5 − 1 = 0.
(b)
Show that (w − 1) (w4 + w3 + w2 + w + 1) = w5 − 1 and deduce that w4 + w3 + w2 + w + 1 = 0.
(c)
Hence show that cos

z1 = 1 i 3

m
2
4
1
 cos   .
5
5
2
and z2 = 1  i  .
n
[12]
16
(a)
Find the modulus and argument of z1 and z2 in terms of m and n, respectively.
(b)
Hence, find the smallest positive integers m and n such that z1 = z2.
[14]