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10

Complex Numbers
Assessment statements
___
√
1
...


Cartesian form z 5 a 1 ib
...

1
...


The complex plane
...
7
De Moivre’s theorem
...

1
...


Introduction
You have already met complex numbers in Chapters 1 and 3
...

Fractals can be generated using
complex numbers
...

The situation, as you already know, is different with quadratic equations
...
The square of any
real number has to be non-negative, i
...

(x 2 > 0 ⇔ x2 1 1 > 1) ⇒ x 2 1 1 > 0 for any choice of a real number x
...
This
forces us to introduce a new set where such a solution is possible
...
However, a question such as 5 1 ? 5 2 is impossible because the
student’s knowledge is restricted to the set of positive integers
...

Also, at early stages an equation such as

x2 5 5
cannot be solved till the student’s knowledge of sets is extended to include irrational
__
√  

...
We extend our number system to
___
 
;
include numbers such as √21 ​ i
...
a number whose square is 21
...
1

___

Numbers such as √21 ​are
​ 
 
not intuitive and many
mathematicians in the past
resisted their introduction, so
they are called imaginary
numbers
...
Euler skilfully employed
them to obtain many
interesting results
...


Complex numbers, sums, products
and quotients
Electronic components like
capacitors are used in AC circuits
...


As you have seen in the introduction, the development of complex
numbers had its origin in the search for methods of solving polynomial
equations
...
However, mathematicians stopped
short of using it for cases where b 2 2 4ac was negative
...

___

___

___

__

√
1
...
​  2k  5 √k   ​  21 ​for any real number k
...

Solution

First we simplify each square root using rule 2
...
For example, to add 5 1 7​  21 ​to 2 2 3​  21 ​
 
 
we combine ‘like’ terms as we do in polynomials:
___

___

___

___

√
√
√
√
(5 1 7​  21 ​ 1 (2 2 3​  21 ​ 5 5 1 2 1 7​  21 ​2 3​  21 ​
 
)
 
)
 
 
___
___
√ 21 ​5 7 1 4​  21 ​
√
 
 

5 (5 1 2) 1 (7 2 3)​ 

Similarly, to multiply these numbers we use the binomial multiplication
procedures:
___

___

___

___

___

√
√
√
√
√
(5 1 7​  21 ​   (2 2 3​  21 ​ 5    2 1 (7​  21 ​   (23​  21 ​ 1 5  (23​  21 ​
 
)
 
) 5
 
)
 
)
 
)
___
√
 
)
1 (7​  21 ​   2
___
___
___
√
√
√
 
)
 
 

5 10 2 21  (​  21 ​ 2 2 15  ​  21 ​1 14  ​  21 ​
___
√
 

5  0 2 21  (21) 1 (215 1 14)​  21 ​
1
___
 

5 31 2 √21 ​
​ 
___

Euler introduced the symbol i for √21 ​
​ 
 

...


Note:  In some cases, especially in engineering sciences, the number i is sometimes
denoted as j
...
For example,
i  3 5 i  2  i 5 21  i 5 2i, and
i  4 5 i  2  i  2 5 (21)  (21) 5 1, and so
i  5 5 i  4  i 5 1  i 5 i, and also
i  6 5 i  4  i  2 5 i  2 5 21; i  7 5 2i, and finally i  8 5 1
...

So, for example i  2122 5 i  2120 1 2 5 i 2 5 21
...

A complex number is a number that can be written in the form a 1 bi where a and
b are real numbers and i  2 5 21
...


We do not define i 5 √21 ​for
​ 
 
a reason
...
We do not mean 23!
i does not belong to this
category since we cannot say
that i is the positive square root
of 21, i
...
i
...
If we do, then
21 5 i  i
...
0, and 21 5 2i  2i
...
Actually
2i is also a square root of 21
because 2i  2i 5 i  2 5 21
...

and write i 5 √ 21 ​
​ 

Notation

It is customary to denote complex numbers with the variable z
...

It is usual to write Re(z) for the real part of z and Im(z) for the imaginary
part
...

Note that both the real and imaginary parts are real numbers!

Algebraic structure of complex numbers

A GDC can be set up to
do basic complex number
operations
...


SCI ENG
FLOAT 0 1 2 3 4 5 6 7 8 9
RADIAN DEGREE
FUNC PAR POL SEQ
CONNECTED DOT
SEQUENTIAL SIMUL
REAL a+bi re^θi
FULL HORIZ G-T
SET CLOCK12/01/08 6:39AM

Gauss’ definition of the complex numbers triggers the following
understanding of the set of complex numbers as an extension to our
number sets in algebra
...
That is, two complex numbers are equal if and only if their real parts
are equal and their imaginary parts are equal
...

For example, if 2 2 ( y 2 2)i 5 x 1 3 1 5i, then x must be 21 and y must
be 23
...

An interesting application of the way equality works is in finding the square roots of
complex numbers without a need for the trigonometric forms developed later in the
chapter
...
Let the square root of z be x 1 yi, then
(x 1 yi )2 5 5 1 12i ⇒ x2 2 y2 1 2xyi 5 5 1 12i ⇒ x2 2 y2 5 5 and
​ 6 ​,
2xy 5 12 ⇒ xy 5 6 ⇒ y 5 __   and when we substitute this value in x2 2 y2 5 5,

(  )

x

6 2
4
​ x​ 
we have x2 2 ​​ __   ​ ​5 5
...
This leads to x 5 62i, that is, the two square roots of 5 1 12i
are 3 1 2i or 23 2 2i
...

Multiplication

(x1, y1)(x2, y2) 5 (x1x2 2 y1y2, x1y2 1 x2y1)
This is equivalent to using the binomial multiplication on (a 1 bi)(c 1 di):
(a 1 bi)  (c 1 di) 5 ac 1 bdi 2 1 adi 1 bci 5 ac 2 bd 1 (ad 1 bc)i
Addition and multiplication of complex numbers inherit most of the
properties of addition and multiplication of real numbers:
z 1 w 5 w 1 z and zw 5 wz  (Commutativity)
z 1 (u 1 v) 5 (z 1 u) 1 v and z(uv) 5 (zu)v  (Associativity)
z (u 1 v) 5 zu 1 zv  (Distributive property)
A number of complex numbers take up unique positions
...

It is therefore normal to identify it with 0
...
So, the real and complex zeros are the
same number
...
This number plays an
important role in multiplication that stems from the following property:
(x, y) (1, 0) 5 (x  1 2 y  0, x  0 1 y  1) 5 (x, y)
432

For complex numbers, (1, 0) behaves like the identity for multiplication for
real numbers
...

The third number of significance is (0, 1)
...
e
...
So, the last result should be
no surprise to us since we know that
i   i 5 21 5 (21, 0)
...
The set of real numbers is
therefore a subset of the set of complex numbers
...
Similarly, pure imaginary numbers are
of the form 0 1 yi 5 (0, y)
...

Notation

So far, we have learned how to represent a complex number in two forms:
(x, y) and x 1 yi
...
)
This last equation justifies why we can write (x, y) 5 x 1 yi
...

a) (4 2 5i) 1 (7 1 8i)
b) (4 2 5i) 2 (7 1 8i)
c) (4 2 5i)(7 1 8i)
Solution

a) (4 2 5i) 1 (7 1 8i) 5 (4 1 7) 1 (25 1 8)i 5 11 1 3i
b) (4 2 5i) 2 (7 1 8i) 5 (4 2 7) 1 (25 2 8)i 5 23 2 13i

(4–5i) (8i)
-
...
5i
Ans Frac
-5 8–1 2i
(4–5i) (7+8i)

68–3i

c) (4 2 5i)(7 1 8i) 5 (4  7 2 (25)  8) 1 (4  8 1 (25)  7)i 5 68 2 3i

Division

Multiplication can be used to perform division of complex numbers
...

433

10

Complex Numbers

Example 4

Find the quotient ______ 
​ 2 1 3i ​ 
...
Hence, using multiplication and the equality of
1 1 2i
complex numbers,

2 1 3i 5 (1 1 2i)(x 1 iy) ⇔ 2 1 3i 5 x 2 2y 1 i(2x 1y)
⇔

{

2 5 x 2 2y

__ ​
__ ​
⇒ x 5 ​ 8 , y 5 ​ 1 
5
5
3 5 2x 1 y

______  5 ​    ​   
Thus, ​ 2 1 3i  __ ​2 __ ​
...
6-
...

c 1 di
With the multiplication as described above:

a 1 bi 5 (cx 2 dy) 1 (dx 1 cy)i
Again by applying the equality of complex numbers property above we get
a system of two equations that can be solved
...


Conjugate
Although the conjugate
notation z * will be used in the
book, in your own work you
can use any notation you feel
comfortable with
...


With every complex number (a 1 bi) we associate another complex
number (a 2 bi) which is called its conjugate
...

​ 

In this book, we will use z  * for the conjugate
...
So the product of a complex number
and its conjugate is always a real number
...

a) z 5 2 1 3i
b) z 5 5i
c) z 5 11
Solution

a) z  * 5 2 2 3i, and (2 1 3i )(2 2 3i) 5 4 2 9i  2 5 4 1 9 5 13
...

c) z  * 5 11, and 11  11 5 121
...

a
ac
_____   _____   _____  __________ ​  1 bd   ______  
​  1 bi​5 ​  1 bi​ ​  2 di   ​  1 2    2 di)​5 ______  ​  2 2 ad​ i
  a
  c
 
​5 (a bi)(c 2 
​1 bc
 
c 1 di c 1 di c 2 di
c   1 d  
c  2 1 d  2
c   1 d  2

Example 6

Find each quotient and write your answer in standard form
...
1061946903–
...
625–
...

(1 1 i )z1 2 iz2 5 23
2z1 1 (1 2 i )z2 5 3 2 3i
435

10

Complex Numbers

Solution

Multiply the first equation by 2, and the second equation by (1 1 i)
...

In the next section, we will add a few more to the list
...

(3) (z1 1 z2) * 5 z1 1 z2
 *
 *

The conjugate of the sum is the sum of conjugates
...
e
...

This result can be generalized
for any non-negative integer
power n, i
...
(z n)* 5 (z *)n and
can be proved by mathematical
induction
...

Now assume (z k)* 5 (z *) k
...

Therefore, (z  k 1 1)* 5 (z *) kz*
5 (z *)k 1 1
...


436

(5) (z1  z2) * 5 z1*  z2*
   

The conjugate of the product is the product of conjugates
...

21

21

Proof

(1) and (2) are obvious
...

(3) is proved by straightforward calculation:

Let z1 5 x1 1 iy1 and z2 5 x2 1 iy2, then
(z1 1 z2) * 5 ((x1 1 iy1) 1 (x2 1 iy2)) * 5 ((x1 1 x2) 1 i (y1 1 y2)) *
5 (x1 1 x2) 2 i (y1 1 y2) 5 (x1 2 iy1) 1 (x2 2 iy2) 5 z1 * 1 z2 *
...

Also (5) is proved by straightforward calculation:
(z1  z2) * 5 ((x1 1 iy1)  (x2 1 iy2)) * 5 ((x1x2 2 y1y2) 1 i (y1x2 1 x1y2)) *
5 (x1x2 2 y1y2) 2 i (y1x2 1 x1y2)
5 (x1 2 iy1)  (x2 2 iy2) 5 z1 *  z2 *

And finally, (6):
(z(z21)) * 5 1 * 5 1
but, (z(z21)) * 5 z *(z21) *, so z  *(z21) * 5 1,
​ 1 
and (z21) * 5 __ ​5 (z  *)21
...

If c is a root of a polynomial equation with real coefficients, then c * is also a
root
...

We give the proof for n 5 3, but the method is general
...


⇒ (ac  ) * 1 (bc  ) * 1 (dc) * 1 e * 5 0

Sum of conjugates theorem
...


3

2

3

2

⇒ (c  *) is a root of P(x) 5 0
...
Find all
other zeros
...
Hence,
using the factor theorem, P(x) 5 (x 2 (1 1 2i))(x 2 (1 2 2i))(x 2 c),
where c is a real number to be found
...
c can either be found by division or by
factoring by trial and error
...


Example 91

1 1 2i is a zero of the polynomial
P(x) 5 x 3 1 (i 2 2)x 2 1 (2i 1 5)x 1 8 1 i
...


1

  Not included in present IB syllabus
...
To find the other zeros, we can perform synthetic
substitution
1

1 1 2i

i22
1 1 2i
21 1 3i

1

2i 1 5
27 1 i
22 1 3i

81i
28 2 i
0

This shows that P(x) 5 (x 2 1 2 2i)(x 2 1 (21 1 3i)x 2 2 1 3i)
...

Note: x2 1 (21 1 3i)x 2 2 1 3i 5 0 can be solved using the quadratic
formula
...

Exercise 10
...

___

___

  1 5 1 √ 24 ​
​ 
 

  2 7 2 √27 ​
​ 
 

___

  3 26
____

____

√  
  4 2​  49 ​

√

  6 2​  ____ 
​ 225  
 ​ ​
16

√
  5 ​  281 ​
 

Perform the following operations and express your answer in the form a 1 bi
...
Find a and b if (2 1 3i )z 5 7 1 i
...
Solve for x and y
...

__

b) Prove that (1 1 i​ √3 ​ 6n 5 82n, where n  Z1
...

 
)
__

__

√  
 
)
34 a) Evaluate (2​  2 ​1 i​ √2 ​ 2
...

 
)
__

__

√  
c) Hence, find (2​  2 ​1 i​ √ 2 ​ 46
...

______


( |z| 5 √​ 2​+ y2​ ​where z = x + iy
...

36 Find the complex number z and write it in the form a 1 bi if z 5 3 1 _______ ​
​  2i  __ 
2 2 i​ √2 ​
 
37 Find the values of the two real numbers x and y such that (x 1 iy)(4 2 7i ) 5 3 1 2i
...

39 Find the complex number z and write it in the form a 1 bi if ______ ​ z ​ 5 2 2 3i
...

41 a) Find the values of the two real numbers x and y such that (x 1 iy)2 5 2 8 1 6i
...

42 If z  C, find all solutions to the equation z3 2 27i 5 0
...


44 Find a polynomial function with integer coefficients and lowest possible degree
__
_,
 
that has ​ 1 ​ 21 and 3 1 i​ √ 2 ​as zeros
...

46 Given that z 5 5 1 2i is a zero of the polynomial f (x) 5 x3 2 7x2 2 x 1 87, find
the other zeros
...


__ 
48 Let z  C
...

z*
439

10

Complex Numbers

49 Given that z 5 (k 1 i )4 where k is a real number, find all values of k such that
a) z is a real number
b) z is purely imaginary
...


51 Solve the system of equations
...
2

iz1 2 (1 1 i )z2 5 3
(2 1 i )z1 1 iz2 5 4

The complex plane

Our definition of complex numbers as ordered pairs of real numbers
enables us to look at them from a different perspective
...
This
correspondence is embodied in the geometric representation of complex
numbers
...
The complex plane has
two axes, the horizontal axis is called the real axis, and the vertical axis is
the imaginary axis
...
The real part is measured along the real axis and
the imaginary part along the imaginary axis
...


z ϭ x ϩ yi

0
440

Real part x

Imaginary part y

imaginary axis

real axis

Let us consider the sum of two complex numbers:
z1 5 x1 1 y1i, and z2 5 x2 1 y2i
As we have defined addition before:
z1 1 z2 5 (x1 1 x2) 1 (y1 1 y2)i
This suggests that we consider complex numbers as vectors; i
...
we regard
the complex number z 5 x 1 iy as a vector in standard form whose
terminal point is the complex number (x, y)
...
So, adding two complex numbers or
subtracting them, or multiplying by a scalar, are similar in both sets
...

Find z1 1 z2 and z1 2 z2
...

The length, norm, of a vector also has a parallel in complex numbers
...

 
|v  | 5 √ x 2 1 y 2 ​
​ 
For complex numbers, the modulus or absolute value
(or magnitude) of the complex number z 5 x 1 yi is
______

|z  | 5 √x 2 1 y 2 ​
​ 

...

z  z* 5 (x 1 iy)(x 2 iy) 5 x 2 1 y 2,
|z|2 5 x 2 1 y 2, and |z*|2 5 x 2 1 y 2
⇒ z  z* 5 |z |2 5 |z*|2
For example:

It follows immediately that since
_________
2
2

z * 5 x 2 yi ⇒ |z *| 5 √x 1 (2y)
​ 

______
 ​5 ​  x 2 1 y 2 ​ then
 
 
,

√

______

2

√
(3 1 4i )(3 2 4i ) 5 9 1 16 5 25 5 (​  32 1 42 ​
)
 

|z *| 5 |z  |
...

a) A 5{z| |z  | 5 3}

b) B 5{z  | |z  | < 3}

Solution

a) A is the set of complex numbers
whose distance from the origin
is 3 units
...


A
|z| ϭ 3

Ϫ3

3

O

B

b) B is the set of complex numbers
whose distance from the origin is
less than or equal to 3
...


|z| р 3

Ϫ3

O

3

Another important property is the following result:
|z1z2| 5 |z1| |z2|
Proof:

_________________________

​ 
   
|z1z2| 5 |(x1x2 2 y1y2) 1 (x1y2 1 x2y1)i| 5 √(x1x2 2 y1y2)2 1 (x1y2 1 x2y1)2 ​
_________________________________________________

5 √ (x1x2)2 2 2x     y1y2 1 (y1y2)2 1 (x1y2)2 1 2x1y2x2y1 1 (x2y1)2 ​
​ 
1x2
____________________________

5 √(x1x2)2 1 (y1y2)2 1 (x1y2)2 1 (x2y1)2 ​
​ 
   
But,

________

________

_________________

|z1| |z2| 5 √x12 1 y12 ​• √x22 1 y22 ​5 √ (x12 1 y12)(x22 1 y22) ​
​ 
 
​ 
  ​ 
  
____________________________

5 √(x1x2)2 1 (y1y2)2 1 (x1y2)2 1 (x2y1)2 ​
​ 
   
And so the result follows
...

Solution
______

________

|(3 1 4i)(5 1 12i)| 5 |3 1 4i| |5 1 12i  | 5 √9 1 16 ​√25 1 144 ​5 5 3 13 5 65,
​ 
 
​ 
 
____________

_____

or |(3 1 4i)(5 1 12i)| 5 |233 1 56i  | 5 √ (233)2   562 ​5 √ 4255 ​5 65
​ 
1
​ 
 

Trigonometric/polar form of a complex number
imaginary axis

r ϭ |z|

Imaginary part y

z ϭ x ϩ yi

θ
0

Real part x

real axis

We know by now that every complex number z 5 x 1 yi can be considered
as an ordered pair (x, y)
...

The trigonometric form uses the modulus of the complex number as its
distance from the origin, r > 0, and u the angle the ‘vector’ makes with the
real axis
...

Clearly x 5 r cos u and y 5 r  sin u  r 5 √x2 1 y   ​ and tan u 5 __ ​
x
Therefore, z 5 x 1 yi 5 r cos u 1 (r sin u)i 5 r(cos u 1 i sin u)
...

Arg(z) is not unique
...

Note:
The trigonometric form is called ‘modulus-argument’ by the IB
...
Also this trigonometric form is abbreviated, for ease of
writing, as follows:
z 5 x 1 yi 5 r(cos u 1 i sin u) 5 r cis u
...
)
443

10

Complex Numbers

Example 14

Write the following numbers in trigonometric form
...

​ 
  ​   
;
​ 1 
1
Hence, by observing the real and
imaginary parts being positive,
we can conclude that the argument
p
...

b) r 5 √ (​  3 ​ 2 1 (21)2 ​5 √4 ​5 2; tan u 5 ___ ​ The real part is positive,
​  √     
)
​   
​ 21 
√ 3 ​
​   
the imaginary part is negative, and the point is therefore in the fourth
quadrant, so u 5 ____ 
 ​
​ 11p
...

__ 
We can also use u 5 2 ​   ​
6

( 

)

 ​
c) r 5 5 and u 5 ___ since it is on the negative side of the imaginary axis
...

__ 
We can also use u 5 2 ​   ​
2
θ ϭ 3π
2
d) r 5 17 and u 5 0
0

( 

)

x

 z 5 17 (cos 0 1 i sin 0)
z ϭ Ϫ5 i

Example 15

Convert each complex number into its rectangular form
...
643 1 6i  0
...
857 1 4
...

The analogy between complex
numbers and vectors stops at
multiplication
...
Then
z1z2 5 (r1(cos u1 1 i sin u1))(r2(cos u2 1 i sin u2)


5 r1r2[(cos u1 cos u2 2 sin u1 sin u2) 1 i (sin u1 cos u2 1 sin u2 cos u1)]
...

Example 16

__

__

√  
√  
Let z1 5 2 1 2i ​  3 ​and z2 5 21 2 i ​  3 ​

...

b) Evaluate z1z2 by using their trigonometric forms and verify that the two
results are the same
...

2
2

Note: You may observe here that multiplying z1 by z2 resulted in a new
number whose magnitude is twice that of z1 and is rotated by an angle
___
...

an angle of __ ​
​   
3
445

10

Complex Numbers

Example 17

__

√  
Let z1 5 22 1 2i and z2 5 3​  3 ​2 3i
...

Solution
__
___ 
____,
√  
 ​
 ​
z1 5 2​  2 ​ cis ​ 3p and z2 5 6 cis ​ 11p then
 
4
6

(  ( 

(  )
))
___ 
√
5 12​  2 ​ cis​  ___ ​ )​5 12​  2 ​ ​  cos ​ 7p ​1 i sin ​ 7p ​ )​
(​ 7p  √ ( ___ 
12
12
12

( 

)

__
__
__
√  
√  
√  
z1z2 5 12​  2 ​ ​ cis​ ___ 1 ​ 11p ​  ​5 12​  2 ​ cis​ ____ ​5 12​  2 ​ cis​ ___ ​1 2p  ​
 ​
 ​ 
 ​ 
​ 3p  ____ 
​ 31p 
​ 7p 
4
12
12
6



__

__
 

 

Note: You can simplify this answer further to get an exact rectangular form
...
e
...

 
 
 
 
conclude that cos ​ 7p ​5 ​ 
4
4
12
12

( 

)

( 

)

This observation gives us a way of using complex number multiplication in
order to find exact values of some trigonometric functions
...


imaginary axis

z ϭ x ϩ yi

Also, z  z* r (cos u 1 i sin u)  r (cos u 2 i sin u)
5

5 r2(cos2 u 1 sin2 u)

5 r2
...
See the
figure opposite
...

Let
z1 5 r1(cos u1 1 i sin u1) and z2 5 r2(cos u2 1 i sin u2)
be two complex numbers written in trigonometric form
...

5 __  ________________________________________________​  ​
​  1 ​​ ​ 
r2
1
Now, using the subtraction formulas for sine and cosine, we have
z
r
__ ​ ​  1 ​
​ z1 5 __  (cos(u1 2 u2)) 1 i(sin(u1 2 u2))]
r2[
2
This formula says: To divide two complex numbers written in trigonometric
form, we divide the moduli and subtract the arguments
...
e
...

1
__
If z 5 r(cos u 1 i sin u) then ​ 1 ​5 __ ​(cos(2u) 1 i sin(2u)) 5 __ ​(cos(u) 2 i sin (u))
​ 1 
r
z ​ r  
Example 18

__

Let z1 5 1 1 i and z2 5 √3 ​2 i
...

1
...
 ​
2
___ 
___ 

...


( 

)

__

__

(  )

__
z
√ 2 ​
√ 2 ​
​   
​   
p
p
__ ​
__  ​ ​    __ 
​ 1 ​  __ 
​   ​ cis​ ​ p  __ 
​   ​ cis​ ​ 5p  ,
​ z1 5 z1  ​ 1  ​5 ( √2 ​ cis ​   ​ )​  ​ __ cis ​   ​  ​5 ___  ( __ ​1 ​ p ​ )​5 ___  ___ ​  ​ or
z2
4
4
2
2
2
12
6
6
2
__
p
__
__
__ 
√ 2 ​ cis ​   ​
​   
z
√  
​   
4  ​ √2 ​ p ____  ​ ​  2 ​ ​ 5p 
__ ​ ​ 
 ​ cis​
 ​ 
​ z1 5 _______  ___  ( __ ​2 ​ 2p ​5 ___  ___ ​  ​
 
2p 5 ​   ​ cis​ ​ 4  
____  2
2
12
6 )
2
2 cis ​   ​
6

(  )

447

10

Complex Numbers
__

__

__

√ 3 ​2 1 1 (​  3 ​1 1)i
√  
z
​   
√ 3 ​1 i
​   
__ ​ ​  1 1 i  ______  _________________ 
__
 ​
d) ​  1  5 ______ ​  ​  __
 
  
 ​5 ​ 

z2

4

√ 3 ​2 i √ 3 ​1 i
​   
​   

Comparing this to part c)
...

 
​   ​5 ​   ​ cos ​ 5p ​⇒ cos ​ 5p ​5 ​   ​  ​  2   ​5 ________ 

4

2

12

__
√ 3 ​1 1
​   
______

Also, ​ 

4

12

__
√ 2 ​
​ 
___ 

___ 
 ​5 ​   ​ sin ​ 5p ​⇒ sin ​ 5p ​5 ​ 
 
 
  ___ 

4

2

12

4

√ 2 ​
​   

__

__
√ 3 ​1 1
​   
______

12

__

√ 6 ​1 √ 2 ​
​    ​   
 ​  ​  2   ​5 ________ 
 ​
 ___  ​ 
  __

...
2

In questions 1–14, write the complex number in polar form with argument u, such
that 0 < u , 2p
...

2

__ 
__ 
__ 
__ 
,
15 z1 5 cos ​ p ​1 i sin ​ p ​ z2 5 cos ​ p ​1 i sin ​ p ​

2

2

3

3

___ 
___,
___ 
___ 
 
 ​
 ​
 ​
 ​
16 z1 5 cos ​ 5p 1 i sin ​ 5p z2 5 cos ​ 7p 1 i sin ​ 7p 

6
6
6
6
__ 
__ 
___ 
___ 
,
 ​
 ​
17 z1 5 cos ​ p ​1 i sin ​ p ​ z2 5 cos ​ 2p 1 i sin ​ 2p 
6
6
3
3
____ 1
____,
___ 
___ 
 
 ​
 ​
18 z1 5 cos ​ 13p  i sin ​ 13p z2 5 cos ​ 5p ​1 i sin ​ 5p ​
12
12
12
12
___ 
___  ,
___ 
___ 
​ 2 
 ​
 ​ 
 ​
 ​ 
19 z1 5 3​ cos ​ 3p 1 i sin ​ 3p  ​ z2 5 __ ​ ​ cos ​ 4p 1 i sin ​ 4p  ​
4
4
3
3
3
__
___ 
___  ,
___ 
___ 
√  
 ​
 ​ 
 ​
 ​ 
20 z1 5 3​  2 ​ ​ cos ​ 5p 1 i sin ​ 5p  ​ z2 5 2​ cos ​ 5p 1 i sin ​ 5p  ​
4
4
3
3

( 

( 

)

( 

)

( 

)

)

21 z1 5 cos 135° 1 i sin 135°, z2 5 cos 90° 1 i sin 90°
22 z1 5 3(cos 120° 1 i sin 120°), z2 5 2(cos 240° 1 i sin 240°)
__

√ 3 ​
​   
​ 5 
​   ​ (
23 z1 5 __ ​ (cos 225° 1 i sin 225°), z2 5 ___ cos 330° 1 i sin 330°)
8
2
__

√  
24 z1 5 3​  2 ​ (cos 315° 1 i sin 315°), z2 5 2(cos 300° 1 i sin 300°)

1, 1,
​ z   ​ z  
In questions 25–30, write z1 and z2 in polar form, and then find the reciprocals __  ​ __  ​
z1
1 2
__  (2p , u , p)
...

_
...

(i) Sketch a diagram to show the points which represent z1 and z2 in the
complex plane
...

32 Use the Argand diagram to show that |z1 1 z2| < |z1| 1 |z2|
...

2
______
______ 
______ 
a) ​  __3   ​  
b) ​  2z  2 ​
c) ​ 3 2 z2 ​
√ 3 ​1 z
31z
31z
​   

34 Find the modulus and argument (amplitude) of each of the complex numbers
__

__

√  
√  
z1 5 2​  3 ​2 2i, z2 5 2 1 2i and z3 5 (2​  3 ​2 2i  )(2 1 2i  )
...

36 Identify, in the complex plane, the set of points that correspond to the following
equations
...

a) |z | < 3
b) |z 2 3i  | > 2


10
...

Let z 5 r (cos u 1 i sin u), now
z 2 5 (r (cos u 1i sin u))(r (cos u 1 i sin u))
5 r 2((cos u cos u 2 sin u sin u) 1 i (sin u cos u 1 cos u sin u))
5 r 2((cos2 u 2 sin2 u) 1 i (2 sin u cos u)) 5 r 2(cos 2u 1 i sin 2u)
...

In general, we obtain the following theorem, named after the French
mathematician A
...

449

10

Complex Numbers

Note: As a matter of fact, de Moivre stated ‘his’ formula only implicitly
...

De Moivre’s theorem
If z 5 r (cos u 1 i sin u) and n is a positive integer, then
z n 5 (r  (cos u 1 i sin u))n 5 r n(cos nu 1 i sin nu)
...


Proof

The proof of this theorem follows as an application of mathematical
induction
...

Basis step:
To prove this formula the basis step must be P(1)
...
]
Inductive step:
Assume that P(k) is true, i
...

z k 5 r k(cos ku 1 i sin ku)
...

So we have to show that z k11 5 r k 1 1(cos(k 1 1)u 1 i sin(k 1 1)u)
...

Note: In fact the theorem is valid for all real numbers n
...

Example 19

Find (1 1 i)6
...

450

__
p
p
__ 
(1 1 i) 5 √2 ​ ​  cos ​   ​1 i sin ​   ​ )​
​   ( __ 
4
4
Now we can apply De Moivre’s theorem
...

(1 1 i)6 5 1 1 6i 1 15i  2 1 20i  3 1 15i  4 1 6i  5 1 i  6
5 1 1 6i 2 15 2 20i 1 15 1 6i 2 1 5 8i
When the powers get larger, we are sure you will appreciate De Moivre!

Applications of De Moivre’s theorem
Several applications of this theorem prove very helpful in dealing with
trigonometric identities and expressions
...

z21 5 r21(cos(2u) 1 i sin(2u)) 5 __ ​(cos u 2 i sin u)
​ 1 
r
Also,
z2n 5 (z21)n 5 (r21(cos(2u) 1 i sin(2u)))n 5 r2n(cos(2nu) 1 i sin(2nu))
...

These relationships are quite helpful in allowing us to write powers of cos u
and sin u in terms of cosines and sines of multiples of u
...

Solution

Starting with
1
​ z
​​
( z 1 __ ​ )​​ ​5 (2 cos u)3
3

and expanding the left-hand side, we get
3
​ z ​ 13 
​ 13 
z3 1 3z 1 __ ​1 __  ​5 8 cos3 u ⇒ z3 1 __  ​1 3​  z 1 __ ​ )​5 8 cos3 u
( ​ 1
z
z 
z 

⇕
⇕

⇒ 2 cos 3u 1 3(2 cos u) 5 8 cos3 u


_​ 
⇒ cos3 u 5 ​ 1 (2 cos 3u 1 3(2 cos u))
8



1
5 _ (cos 3u 1 3 cos u)
​  ​ 
4

451

10

Complex Numbers

Example 21

Simplify the following expression:
(cos 6u 1 i sin 6u)6(cos 3u 1 i sin 3u)
 
   
  
​  _____________________________​
cos 4u 1 i sin 4u
Solution
6

(cos 6u 1 i sin 6u) (cos 3u 1 i sin 3u)
______________________________​
 
​ 
   
  

cos 4u 1 i sin 4u
(cos u 1 i sin u)6(cos u 1 i sin u)3
___________________________
5 ​ 
   
    ​
(cos u 1 i sin u)4
Using the laws of exponents, we have

(cos u 1 i sin u)6(cos u 1 i sin u)3
___________________________
   
    ​ 5 (cos u 1 i sin u)5
​ 
(cos u 1 i sin u)4
5 cos 5u 1 i sin 5u
...

An nth root of a given number z is a number w that satisfies the following
relation
wn 5 z
...

__

__

√  
This is also because w 10 5 (2​  3 ​1 i  )10 5 512 1 512i √3 ​
​   

...

Let w 5 s (cos a 1 i sin a) be an nth root of z 5 r (cos u 1 i sin u)
...
e
...


Also,
cos na 5 cos u and sin na 5 sin u
...

452

This leads to

u   ​5
a 5 _______  __ 1 ____ k 5 0, 1, 2, 3,
...

​  1n2kp  ​ u ​ ​ 2kp​
 
n
n  ;
Notice that we stop the values of k at n 2 1
...

nth roots of a complex number
Let z 5 r (cos u 1 i sin u) and let n be a positive integer, then z has n distinct nth roots

(  ( 

))

( 

)

n _
u
u
√  
​ n ​ ​ 2kp 
​ n ​ ​ 2kp 
zk 5 ​   r ​ ​ cos​ __ 1 ____ ​ 1 i sin​ __ 1 ____ ​  ​
n ​ 
n ​ 

where k 5 1, 2, 3, …, n 2 1
...
Thus all these roots lie
r
n

1
__

_

√
on a circle in the complex plane whose radius is ​   r ​ 5 ​ ​ ​n  ​​
...

consecutive roots differ by ​  n ​,
n

Example 22

Find the cube roots of z 5 28 1 8i
...

(6   
...


(  ( 

( 

)

))

( 

6 __
6 __
p
w4 5 2​√2 ​​ cos​ __ ​1 ​ 6p  ​1 i sin​ __ ​1 ​ 6p  ​  ​5 2​√2 ​​ cos​  __ ​1 2p  ​1 i sin​  __ ​1 2p  ​  ​
 ​ 
 ​ 
   
​ p  ___ 
​    ___ 
   
( ​ p  )
( ​ p  )
4
4
4
4
3
3
6 __
p ​  ​1 i sin​  __ ​  ​  ​5 w
p 
5 2​√2 ​​ cos​  __ )
   
( ​ 4
( ​ 4 )
1

( 

)

)

Also, if you raise any of the roots to the third power, you will eventually get
z; for example,

(  ))
(  )) ]
(  (  )
[  (  (  )
√ (
5 8​  2 ​​  cos​  ____ ​1 i sin​  ____ ​ )​5 8​  2 ​​  cos​  ___  ​1 i sin​  ___  ​ )​5 z
 ​ 
 ​ 
(​ 34p ​  )
(​ 11p )
(​ 11p ) √ ( (​ 34p ​  )
4
4

3
__
6 __
√  
 ​ 
 ​ 
 ​ 
 ​ 
   
​ 11p 
​ 11p 
​ 33p 
​ 33p 
(w2)3 5 ​​ 2​√2 ​​ cos​ ____ ​1 i sin​ ____ ​  ​  ​​ ​5 8​  2 ​​ cos​ ____ ​1 i sin​ ____ ​  ​
12
12
12
12
__



__
 

 

Example 23

Find the six sixth roots of z 5 264 and graph these roots in the complex
plane
...
So the roots are

(  ( 

)

))

( 

u
u
​ n ​ ​ 2kp​ 
​ n​ ​ 2kp​ 
w 5 s​ cos​ __ 1 ____ ​1 i sin​ __ 1 ____ ​  ​
n   
n   

(  ( 
(  ( 

( 

)

))

__
6
p 
 ​ 
​ p  ____ 
​ p  ____ 
5 √64 ​​ cos​ __ ​1 ​  2k ​  ​1 i sin​ __ ​1 ​ 2kp ​  ​
​   
6
6
6
6
 ​ 
 ​ 
​ p  ___ 
​ p  ___  ;
5 2​ cos​ __ ​1 ​ kp  ​1 i sin​ __ ​1 ​ kp  ​  ​ k 5 0, 1, 2, 3, 4, 5
3
3
6
6

( 

w3

w2

Ϫ3

Ϫ2

1

w4

Ϫ1

( 
( 

0
Ϫ1
Ϫ2 w
5

π
6

1

2
w6

3

x

)

)

)
( 
))
(  ( 
(  ))
(  (  )
w 5 2​  cos​  __ ​1 ​ 3p  ​1 i sin​  __ ​1 ​ 3p  ​ )​
 ​ 
(​ p  ___ )
( (​ p  ___ ​  )
3
3
6
6
5 2​  cos​  ​ 7p  )​ i sin​  ___  ​ )​
( (___ ​  1 (​ 76p ​  )
6
w 5 2​  cos​  __ ​1 ​ 4p  ​1 i sin​  __ ​1 ​ 4p  ​ )​
 ​ 
(​ p  ___ )
( (​ p  ___ ​  )
3
3
6
6
5 2​  cos​  ___  ​1 i sin​  ___  ​ )​
(​ 32p ​  )
( (​ 32p ​  )
 ​ 
w 5 2​  cos​  __ ​1 ​ 5p  ​1 i sin​  __ ​1 ​ 5p  ​ )​
(​ p  ___ )
( (​ p  ___ ​  )
3
3
6
6
 ​ 
 ​ 
5 2​  cos​  ____ ​1 i sin​  ​ 11p ​ )​
(____ )
( (​ 11p )
6
6

w3 5 2​ cos​ __ ​1 ​ 2p  ​1 i sin​ __ ​1 ​ 2p  ​  ​
 ​ 
 ​ 
​ p  ___ 
​ p  ___ 
3
3
6
6
 ​ 
 ​ 
​ 5p 
​ 5p 
5 2​ cos​ ___  ​1 i sin​ ___  ​  ​
6
6
4

5

6

454

)

w2 5 2​ cos​  __ ​1 ​ p ​ )​1 i sin​  __ ​1 ​ p ​ )​  ​
( ​ p  __ 
( ​ p  __ 
3
3
6
6
p 
p 
__ ​  ​1 i sin​  __ ​  ​  ​
5 2​ cos​  ​  )
(2
( ​ 2 )

w1
π
3

))

w1 5 2​ cos​  __ ​ )​1 i sin​  __ ​ )​  ​
( ​ p 
( ​ p 
6
6

y
2

( 

)

nth roots of unity

The rules we established can be applied to finding the nth roots of 1
(unity)
...
We can write it as
1 5 1(cos 0 1 i sin 0)
...
, n 2 1
n 
n 

_
n
u
u
   
 
​
​ n ​ ​ 2kp​ 
​ n ​ ​ 2kp​ 
zk 5 ​√r  cos​ __ 1 ____ ​1 i sin​ __ 1 ____ ​  ​
n   
n   
n

__
 

Or in degrees,

( 

)

( 

)

360k ​ 
​  n   
​ 360k  ;
zk 5 cos​ ____ ​1 i sin​ ____ ​ k 5 0, 1, 2,
...

Solution

a) Here k 5 2, and therefore the two roots are

(  )
(  )
(  )
(  )
(  )
(  )

360k 
 ​ 
​   ​ 
​ 360k  ;
zk 5 cos​ ____ ​1 i sin​ ____ ​ k 5 0, 1
2
2
0
​   
​ 0 
z0 5 cos​ __ ​  ​1 i sin​ __ ​  ​5 1
2
2
360 ​1 i sin​ ___ ​5 cos 180 1 i sin 180 5 21
 ​ 
​   ​ 
​ 360 
z1 5 cos​ ___ 
2
2
b) Here k 5 3, and the three roots are

(  )
(  )
(  )
(  )
(  )
(  )
(  )
(  )

2k  ​ 
 ​ 
​  p 
​ 2kp  ;
zk 5 cos​ ____ ​1 i sin​ ____ ​ k 5 0, 1, 2, 3
3
3
0
​   ​ 
​ 0 
z0 5 cos​ __  ​1 i sin​ __ ​  ​5 1
3
3
__
√
2p  ​1 i sin​ ___  ​5 2 ​   ​1 i  ​ ​  3 ​
2p 
1
___ 
__ 
___ 
 ​
​   ​ 
 
z1 5 cos​ ​   ​ 
3
3
2
2
__
√ 3 ​
​ 
4  ​ 
1
__ 
___ 
 ​ 
​  p 
​ 4p 
 
z2 5 cos​ ___  ​1 i sin​ ___  ​5 2 ​   ​2 i  ​   ​
3
3
2
2

Euler’s formula
The material in this part depends on work that you will do in the Analysis
option
...

455

10

Complex Numbers

In the options section on infinite series, we have the following results
...
5
​ x    ​ x    ​ x   

∞

2n 1 1

​  x 
 
∑(21)n ________ ​ 
(2n 1 1)!
0
∞

2
4
6
2n
​  x    
cos x 5 1 2 __ ​1 __ ​2 __ ​1
...
5
x

∞

n

x 
∑ ​ __ ​ 
n!
0

Now if you add
2
3
4
5
6
7
sin x 1 cos x 5 1 1 x 2 __ ​2 __ ​1 __ ​1 __ ​2 __ ​2 __ ​1
...

Look at i, i 2, i 3, i  4, i 5, i  6, i 7, i 8,
...

This is known as Euler’s formula
...

Solution

(r (cos u 1 i sin u))n 5 (re iu)n 5 r ne inu
5 r n(cos nu 1 i sin nu)
Example 27

Find the real and imaginary parts of the complex numbers:
p
__

a) z 5 3​  i  ​ 6  ​​
e​

b) z 5 7e 2i

Solution

__

√
3​  3 ​
p,
__  ____ 
a) Since |z  | 5 3 and arg(z) 5 __ ​ Re(z) 5 3 cos ​ p ​5 ​   ​and
​   
 
 
2
6
6
p ​5 ​ 3 
...

Example 28

Express z 5 5 1 5i in exponential form
...

|z  | 5 5​  2 ​and tan u 5 __ ​5 1 ⇒ u 5 __ ​ therefore z 5 5​  2 ​​  i  ​ 4  ​​
​ 5 
​   
 
5
4

Example 29

Evaluate (5 1 5i)6 and express your answer in rectangular form
...
From the example above, z 5 5​√2 ​ ​ i  ​ 4  ​​ hence,
   e  ;
​
z 6 5 ( 5​√2 ​ ​ i  ​ 4  ​​ )​​ ​ (5​  2 ​ 6​  i  ​ 4  ​​ 3 6 5 125 000 ​ i ​ 2  ​​5 2125 000i
...

​   
)
( __ 
4
4
4
4

)

Example 30

Simplify the following expression:
(cos 6u 1 i sin 6u)(cos 3u 1 i sin 3u)
_____________________________​
 
   
  
​ 
cos 4u 1 i sin 4u

Solution

(cos 6u 1 i sin 6u)(cos 3u 1 i sin 3u) e    e 
_____________________________​5 _______  e  u 5 cos 5u 1 i sin 5u
 
​ 
   
  
​  4iu    5i
​5
6iu

cos 4u 1 i sin 4u

3iu

e 

457

10

Complex Numbers

Example 31

Use Euler’s formula to find the cube roots of i
...


Exercise 10
...

2p
__

  1 z 5 4​ 2i ​  3   ​​
e​

  2 z 5 3e2pi

  3 z 5 3e0
...

__

  7 2 1 2i
__

√  
  8 ​  3 ​1i
__

__

√  
  9 ​  6 ​2 i​ √2 ​
 

10 2 2 2i​ √3 ​
 

11 23 1 3i

12 4i

__

√  
13 23​  3 ​2 3i

14 i(3 1 3i )

15 p

16 ei

In questions 17–25, find each complex number
...

__

√  
18 (​  3 ​2 i)6

17 (1 1 i  )10
__

19 (3 1 3i​ √3 ​ 9
 
)
__
__

20 (2 2 2i  )12

__
__

√  
21 (​  3 ​2 i​ √3 ​ 8
 
)

22 (23 1 3i  )7
__

√  
23 (​  3 ​2 i​ √3 ​ 28
 
)

√  
24 (23​  3 ​2 3i  )27

__

√  
25 2(​  3 ​1 i  )7

In questions 26–30, find each root and graph them in the complex plane
...

__

27 The cube roots of 4 1 4i​ √ 3 ​
 

...

29 The sixth roots of i
...

458

In questions 31–36, solve each equation
...

37 (cos(9b) 1 i sin(9b))(cos(5b) 2 i sin(5b))
(cos(6b) 1 i sin(6b))(cos(4b) 1 i sin(4b))
________________________________
       
 ​
38 ​ 
(cos(3b)) 1 i sin(3b))
1
_

)
39 (cos(9b) 1 i sin(9b)​ ​ ​3 ​​
__________________

n
40 ​√(cos(2nb) 1 i sin(2nb)) ​
 
  

41 Use e iu to prove that cos(a 1 b) 5 cos a cos b 2 sin a sin b
...

43 Use De Moivre’s theorem to show that cos 5a 5 16 cos5 a 2 20 cos3 a 1 5 cos a
...

​ 1
8
45 Let z 5 cos 2a 1 i sin 2a
...

​ 1 
z
z
b) Find an expression for cos 2na and sin 2na in terms of z
...
Simplify (1 1 3v)(1 1 3v2)
...

b) Simplify (1 1 b)(1 1 b2 1 b3)
...

48 a) Show that the fifth roots of unity can be written as 1, a, a2, a3 and a4
...

c) Show that 1 1 a 1 a2 1 a3 1 a4 5 0
...

​   
)
​   
)
50 Given that z 5 (2a 1 3i )3, and a  1, find the values of a such that arg z 5 135°
...
Find the values of x and y if (1 2 i )z 5 1 2 3i
...
Evaluate:
a) 1 1 v 1 v2
b) (vx 1 v2y)(vy 1 v2x)
  3 a) Evaluate (1 1 i )2
...

c) Hence or otherwise, find (1 1 i )32
...


2
p
p
...

z2
12
12
z
__ ​
c) Find the value of ​  1 in the form a 1 bi, where a and b are to be determined exactly
z2
___ 
___ 
in radical (surd) form
...

12
12

( 

( 

)

p
p
p
p
__ 
__ 
__ 
__ 
...

Express ​​ ​   ​ 
z2

(  )

  6 If z is a complex number and |z 1 16| 5 4|z 1 1|, find the value of |z |
...

  8 Given that z 5 (b 1 i )2, where b is real and positive, find the exact value of b when
arg z 5 60°
...

  9 The complex number z satisfies i (z 1 2) 5 1 2 2z, where i 5 ​  21 ​ Write z in the
form z 5 a 1 bi, where a and b are real numbers
...

b) Find the zeros of z  5 2 1, giving your answers in the form
r (cos u 1 i sin u ) where r
...

4
c) Express z   1 z  3 1 z  2 1 z 1 1 as a product of two real quadratic factors
...

b) The cube root of 8i which lies in the first quadrant is denoted by z
...


( 

) ( 

)

p
p
p
p
__ 
__  2
__ 
__  3
​​ cos ​   ​ 2 i sin ​   ​  ​​ ​​​ cos ​   ​ 1 i sin ​   ​  ​​ ​
4
4
3
3
_____________________________
12 Consider the complex number z 5 ​ 
        p 4 ​
...

(ii) Find the argument of z, giving your answer in radians
...
e
...

c) Simplify (1 1 2z)(2 1 z 2), expressing your answer in the form a 1 bi, where a and
b are exact real numbers
...

1–i
Express z in the form x 1 i y where x, y  Z
...

b) The complex number z is defined by z 5 cos u 1 i sin u
...

(ii) Deduce that z  n 1 z2n 5 2 cos nu
...

__​ 
(ii) Hence, show that cos5 u 5 ​ 1  (a cos 5u 1 b cos 3u 1 c cos u), where a, b and
16
c are positive integers to be found
...
Find p and q
...


( 

)

2  ​ 1
2  ​ 
___
___
(i) Show that z1 5 2​ cos ​  p  i sin ​  p  ​ is one of the complex roots of this equation
...

1 1 1
1
(iii) Plot the points that represent z1, z  2, z  3, z   4 and z  5 in the complex plane
...
Give a full geometric description of the two transformations
...

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
...

__ 
(ii) Show that arg(z1) 5 ​   ​
6
(iii) Find arg(z2)
...

2i
18 Given that (a 1 i )(2 2 bi ) 5 7 2 i, find the value of a and of b, where a, b  Z
...

a) Using De Moivre’s theorem show that
__ 
z  n 1 ​ 1n  ​5 2 cos nu
...

8
1
1
_​ 
_​ 
20 Consider the complex geometric series e iu 1 ​   e2iu 1 ​   e3iu 1 …
4
2
a) Find an expression for z, the common ratio of this series
...

c) Write down an expression for the sum to infinity of this series
...

(ii) Hence, show that

4 cos u 2 2 
1
1
_​ 
_​ 
_________ 
...
Two of the roots of P (z  ) 5 0 are
22 and (23 1 2i )
...

__

√  
,
22 Given that |z | 5 2​  5 ​ find the complex number z that satisfies the equation

15​
___ ​2 ​    5 1 2 8i
...
Denote its two roots by z1 and z2 and
express them in exponential form with z1 in the first quadrant
...

z  
2
c) Show that z   4 5 z   4
...

n
e) For what values of n is z 1 real?

2  ​ 1
2  ​ is
___
___
25 a) Show that z 5 cos ​  p  i sin ​  p  a root of the equation x7 2 1 5 0
...

2  ​ 1
4  ​ 1
6  ​ 5 _​
1
___
___
___
c) Show that cos ​  p  cos ​  p  cos ​  p  2 ​   

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