Download Chapter 10 - Complex Numbers

10
Complex Numbers
Assessment statements
___
1.5
Complex numbers: the number i 5 ​ √ 21 ​ 
; the term’s real part,
imaginary part, conjugate, modulus and argument.
Cartesian form z 5 a 1 ib.
Sums, products and quotients of complex numbers.
1.6
Modulus–argument (polar) form z 5 r (cosu 1 i sinu) = rcis(u) = reiu.
The complex plane.
1.7
De Moivre’s theorem.
Powers and roots of a complex number.
1.8
Conjugate roots of polynomial equations with real coefficients.
Introduction
You have already met complex numbers in Chapters 1 and 3. This chapter
will broaden your understanding to include trigonometric representation
of complex numbers and some applications.
Fractals can be generated using
complex numbers.
Solving a linear equation of the form
ax 1 b 5 0, with a  0
is a straightforward procedure if we are using the set of real numbers.
The situation, as you already know, is different with quadratic equations.
For example, as you have seen in Chapter 3, solving the quadratic equation
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x 2 1 1 5 0 over the set of real numbers is not possible. The square of any
real number has to be non-negative, i.e.
(x 2 > 0 ⇔ x2 1 1 > 1) ⇒ x 2 1 1 > 0 for any choice of a real number x.
This means that x 2 1 1 5 0 is impossible for every real number x. This
forces us to introduce a new set where such a solution is possible.
The situation with finding a solution to x2 1 1 5 0 is analogous to the following
scenario: For a child in the first or second grade, a question such as 5 1 ? 5 9 is
manageable. However, a question such as 5 1 ? 5 2 is impossible because the
student’s knowledge is restricted to the set of positive integers.
However, at a later stage when the same student is faced with the same question,
he/she can solve it because their scope has been extended to include negative
numbers too.
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
numbers where he/she can recognize numbers such as x 5 6​ √5 ​ .
The situation is much the same
for x2 1 1 5 0. We extend our number system to
___
include numbers such as √
​  21 ​ ; i.e. a number whose square is 21.
10.1
___
Numbers such as √
​  21 ​ are
not intuitive and many
mathematicians in the past
resisted their introduction, so
they are called imaginary
numbers.
Thanks to Euler’s (1707–1783)
seminal work on imaginary
numbers, they now feature
prominently in the number
system. Euler skilfully employed
them to obtain many
interesting results. Later, Gauss
(1777–1855) represented them
as points in the plane and
renamed them as complex
numbers, using them to
obtain various significant
results in number theory.
Complex numbers, sums, products
and quotients
Electronic components like
capacitors are used in AC circuits.
Their effects are represented using
complex numbers.
As you have seen in the introduction, the development of complex
numbers had its origin in the search for methods of solving polynomial
equations. The quadratic formula
________
√
​  b 2 2 4ac  
​
​ 
   
​ 
x 5 ___
​ 2b ​ 6 _________
2a
2a
had been used earlier than the 16th century to solve quadratic equations –
in more primitive notations, of course. However, mathematicians stopped
short of using it for cases where b 2 2 4ac was negative. The use of the
formula in cases where b 2 2 4ac is negative depends on two principles (in
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10
Complex Numbers
addition to the other principles inherent in the set of real numbers, such as
associativity and commutativity of multiplication).
___
___
___
__
1.​ √21 ​   ​ √21 ​ 5 21
___
2.​ √2k  ​5 √
​  k  ​  ​ √21 ​ for any real number k . 0
Example 1
____
____
Multiply √
​  236 ​   ​ √249 ​ 
.
Solution
First we simplify each square root using rule 2.
____
___
___
___
____
___
___
___
√
​  236 ​ 5 √
​  36 ​   ​ √21 ​ 5 6  ​ √21 ​ 
√
​  249 ​ 5 √
​  49 ​   ​ √21 ​ 5 7  ​ √21 ​ 
And hence using rule 1 with the other obvious rules:
____
____
___
___
___
___
√
​  236 ​   ​ √ 249 ​ 5 6  ​ √21 ​   7  ​ √21 ​ 5 42  ​ √21 ​   ​ √21 ​ 5 242
To deal with the quadratic formula expressions that consist of
combinations of real numbers and square roots of negative numbers, we
can apply the rules of binomials to numbers of the form
___
a 1 b ​ √ 21 ​ 
___
___
where a and b are real numbers. 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 ​ 
___
___
5 (5 1 2) 1 (7 2 3)​ √21 ​ 5 7 1 4​ √ 21 ​ 
Similarly, to multiply these numbers we use the binomial multiplication
procedures:
___
___
___
___
___
(5 1 7​ √21 ​ )  (2 2 3​ √21 ​ ) 5 5  2 1___
(7​ √21 ​ )  (23​ √21 ​ ) 1 5  (23​ √21 ​ )
1 (7​ √21 ​ )  2
___
___
___
5 10 2 21  (​ √ 21 ​ )2 2 15  ​ √ 21 ​ 1 14  ​ √21 ​ 
___
5 10 2 21  (21) 1 (215 1 14)​ √21 ​ 
___
5 31 2 √
​  21 ​ 
___
Euler introduced the symbol i for √
​  21 ​ .
A pure imaginary number is a number of the form ki, where k is a real number and i,
the imaginary unit, is defined by i 2 5 21.
Note:  In some cases, especially in engineering sciences, the number i is sometimes
denoted as j.
Note:  With this definition of i, a few interesting results are immediately
apparent. 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.
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This leads you to be able to evaluate any positive integer power of i using
the following property:
i 4n 1 k 5 i k, k 5 0, 1, 2, 3.
So, for example i 2122 5 i 2120 1 2 5 i 2 5 21.
___
Example 2
Simplify
____
____
____
a)​ √236 ​ 1 √
​  249 ​ 
____
b)​ √236 ​   ​ √249 ​ 
Solution
____
____
___
___
___
___
a)​ √236 ​ 1 √
​  249 ​ 5 √
​  36 ​ ​  √21 ​ 1 √
​  49 ​ ​  √21 ​ 
5 6i 1 7i 5 13i
____
____
b)​ √236 ​   ​ √249 ​ 5 6i  7i 5 42i 2
5 42(21) 5 242
Gauss introduced the idea of complex numbers by giving them the
following definition.
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. a is called the real part of the number and b is the
imaginary part.
We do not define i 5 √
​  21 ​ for
a reason. It is the convention
in mathematics
that when we
__
write √
​  9 ​ then we mean the
non-negative square root of 9,
namely 3. 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.e. i . 0. If we do, then
21 5 i  i . 0, which is false,
and if we say i , 0, then
2i . 0, and 21 5 2i  2i .
0, which is also false. Actually
2i is also a square root of 21
because 2i  2i 5 i 2 5 21.
With this in mind, we can use
a ‘convention’ which calls i the
principal square
root of 21
___
and write i 5 √
​  21 ​ .
Notation
It is customary to denote complex numbers with the variable z.
z 5 5 1 7i is the complex number with real part 5 and imaginary part 7
and z 5 2 2 3i has 2 as real part and 23 as imaginary.
It is usual to write Re(z) for the real part of z and Im(z) for the imaginary
part. So, Re(2 1 3i ) 5 2 and Im(2 1 3i ) 5 3.
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. For example, if you
have a TI-84 Plus, the set up is
as follows.
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.
The set of complex numbers C is the set of ordered pairs of real numbers
C 5 {z 5 (x, y): x, y  }, with the following additional structure:
Equality
Two complex numbers z1 5 (x1, y1) and z2 5 (x2, y2) are equal if their
corresponding components are equal: (x1, y1) 5 (x2, y2) if x1 5 x2 and
y1 5 y2. That is, two complex numbers are equal if and only if their real parts
are equal and their imaginary parts are equal.
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10
Complex Numbers
This is equivalent to saying: a 1 bi 5 c 1 di ⇔ a 5 c and b 5 d.
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. Explain why.
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.
Find the square root(s) of z 5 5 1 12i. 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  ​, and when we substitute this value in x2 2 y2 5 5,
2xy 5 12 ⇒ xy 5 6 ⇒ y 5 __
(  )
x
6  ​  2​ ​5 5. This simplifies to x 4 2 5x2 2 36 5 0 which yields x2 5 24
​ x
we have x2 2 ​​ __
or x2 5 9, ⇒ x 5 63. This leads to x 5 62i, that is, the two square roots of 5 1 12i
are 3 1 2i or 23 2 2i.
(3+2i)2
(-3–2i)2
5+12i
5+12i
Addition and subtraction for complex numbers are defined as follows:
Addition
(x1, y1) 1 (x2, y2) 5 (x1 1 x2, y2 1 y2)
This is equivalent to saying: (a 1 bi) 1 (c 1 di) 5 (a 1 c) 1 (b 1 d)i.
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. For example, the
number (0, 0) has the properties of 0:
(x, y) 1 (0, 0) 5 (x, y) and (x, y)(0, 0) 5 (0, 0).
It is therefore normal to identify it with 0. The symbol is exactly the same
symbol used to identify the ‘real’ 0. So, the real and complex zeros are the
same number.
Another complex number of significance is (1, 0). 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. Again, it is normal to write (1, 0) 5 1.
The third number of significance is (0, 1). It has the notable characteristic
of having a negative square, i.e.
(0, 1)(0, 1) 5 (0  0 2 1  1, 0  1 1 1  0) 5 (21, 0)
Using the definition above, (0, 1) 5 0 1 1i 5 i. So, the last result should be
no surprise to us since we know that
i  i 5 21 5 (21, 0).
Since (x, y) represents the complex number x 1 yi, then every real
number x can be written as x 1 0i 5 (x, 0). The set of real numbers is
therefore a subset of the set of complex numbers. They are the complex
numbers whose imaginary part is 0. Similarly, pure imaginary numbers are
of the form 0 1 yi 5 (0, y). They are the complex numbers whose real part
is 0.
Notation
So far, we have learned how to represent a complex number in two forms:
(x, y) and x 1 yi.
Now, from the properties above
(x, y) 5 (x, 0) 1 (0, y) 5 (x, 0) 1 (y, 0)(0, 1)
(Check the truth of this equation.)
This last equation justifies why we can write (x, y) 5 x 1 yi.
Example 3
Simplify each expression.
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)
-.625–.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.
a 1 bi  
​, involves finding a complex
The division of two complex numbers, ​ ______
c 1 di
number (x 1 yi) satisfying ______
​ a 1 bi  
​5 x 1 yi; hence, it is sufficient to find
c 1 di
the unknowns x and y.
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10
Complex Numbers
Example 4
 ​ .
Find the quotient ______
​ 2 1 3i 
1 1 2i
Solution
2 1 3i 
 ​5 x 1 iy. Hence, using multiplication and the equality of
Let ​ ______
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
8  ​, y 5 ​ __
1  ​
⇒ x 5 ​ __
5
5
3 5 2x 1 y
2 1 3i 
Thus, ​ ______
 ​5 __
​ 8 ​ 2 __
​ 1 ​ i.
1 1 2i 5 5
(2+3i) (1+2i)
1.6-.2i
Ans Frac
8 5–1 5i
a 1 bi  
​5 x 1 yi ⇔ a 1 bi 5 (x 1 yi)(c 1 di).
Now, in general, ​ ______
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.
{
cx 2 dy 5 a
bc 2 ad ​ 
⇒ x 5 _______
​ ac2 1 bd2 ​ 
; y 5 ​ _______
c 1
d c 2 1 d 2
dx 1 cy 5 b
The denominator c 2 1 d 2 resulted from multiplying c 1 di by c 2 di ,
which is its conjugate.
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. You just
need to understand that the IB
questions use this one.
With every complex number (a 1 bi) we associate another complex
number (a 2 bi) which is called its conjugate. The conjugate of number z
is most often denoted with a bar over it, sometimes with an asterisk to the
right of it, occasionally with an apostrophe and even less often with the
plain symbol Conj as in
_
z​
​  5 z * 5 z9 5 Conj(z).
In this book, we will use z * for the conjugate.
The importance of the conjugate stems from the following property
(a 1 bi )(a 2 bi ) 5 a2 2 b 2i 2 5 a2 1 b 2
which is a non-negative real number. So the product of a complex number
and its conjugate is always a real number.
434
Example 5
Find the conjugate of z and verify the property mentioned above.
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.
b) z * 5 25i, and (5i )(25i ) 5 25i2 5 (25)(21) 5 5.
c) z * 5 11, and 11  11 5 121.
So, the method used in dividing two complex numbers can be achieved by
multiplying the quotient by a fraction whose numerator and denominator are the
conjugate c 2 di.
2 di) ______
_____
​ a 1 bi  ​ 5 _____
​ a 1 bi  ​   _____
​ c 2 di  ​ 5 __________
​ (a 1 2bi)(c   
  ​ 
5 ​ ac2 1 bd2  
​ 
1 ______
​ bc2 2 ad2  
​  
i
c 1 di c 1 di c 2 di
c 1 d 2
c 1 d c 1 d Example 6
Find each quotient and write your answer in standard form.
4 2 5i 
 ​
a)​ ______
7 1 8i
4 2  5i 
b)​ ______
​ 
8i
4 2 ​
5i 
c)​ ______
 
7
Solution
28 2 40 1 (232 2 35)i
7 2 8i 
4 2 5i 
12
​  67  ​ i
 ​5 ______
​ 4 2 5i 
 ​  ​ ______
 ​5 ____________________
​ 
 ​
a)​ ______
  
   5 2 ​ ___  ​ 2 ___
113 113
49 1 64
7 1 8i 7 1 8i 7 2 8i
4 2  5i 
4 2  5i 
28i  ​5 _________
40 
5 ​ 2 __
b)​ ______
​ 1 ​ i
​ 5 ​ ______
​   ​ ____
​ 232i 2 ​
5 2 ​ __
 
8 2
64
8i
8i
28i
4 2 ​
5i 
4 ​ 5 ​ i
c)​ ______
 5 ​ __  ​2 __
7
7 7
(4–5i) (7+8i)
-.1061946903–.5…
Ans Frac
-12 113–67 113i
(4–5i) (8i)
-.625–.5i
Ans Frac
-5 8–1 2i
Example 7
Solve the system of equations and express your answer in Cartesian form.
(1 1 i )z1 2 iz2 5 23
2z1 1 (1 2 i )z2 5 3 2 3i
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10
Complex Numbers
Solution
Multiply the first equation by 2, and the second equation by (1 1 i).
2(1 1 i )z1 2 2iz25 26 2(1 1 i )z1 1 (1 1 i)(1 2 i )z25 (1 1 i )(3 2 3i )
2(1 1 i )z1 1 2z25 6 (1)
(2)
By subtracting (2) from (1), we get
(22 2 2i )z25 212
​5 3 2 3i
z2 5 _______
​  212   
22 2 2i
23 1 i(3 2 3i) __
z15 _____________
​ 3  ​
 
​5 ​ 3 ​ 1 __
​ 
  
 
2 2i
11i
And hence
Properties of conjugates
Here is a theorem that lists some of the important properties of conjugates.
In the next section, we will add a few more to the list.
Theorem
Let z, z1 and z2 be complex numbers, then
(1)(z *) * 5 z
(2)z * 5 z if and only if z is real.
(3)(z1 1 z2) * 5 z1 * 1 z2 *
The conjugate of the sum is the sum of conjugates.
(4)(2z) * 5 2z *
The product can be extended
to powers of complex numbers,
i.e.
(z2)* 5 (z  z)* 5 z*  z* 5 (z*)2.
This result can be generalized
for any non-negative integer
power n, i.e. (z n)* 5 (z *)n and
can be proved by mathematical
induction.
The basis case, when n 5 0, is
obviously true:
(z 0)* 5 1 5 (z *)0.
Now assume (z k)* 5 (z *) k.
(z k 1 1)* 5 (z kz)* 5 (z k)*z*
5 (z *) kz * (using the product
rule).
Therefore, (z k 1 1)* 5 (z *) kz*
5 (z *)k 1 1.
So, since if the statement is
true for n 5 k, it is also true for
n 5 k 1 1, then by the principle
of mathematical induction it is
true for all n  0.
436
(5)(z1  z2) * 5 z1 *  z2 *
The conjugate of the product is the product of conjugates.
(6)(z ) * 5 (z *) , if z  0.
21
21
Proof
(1) and (2) are obvious. For (1), ((a 1 bi) *) * 5 (a 2 bi) * 5 a 1 bi, and
for (2), a 2 bi 5 a 1 bi ⇒ 2bi 5 0 ⇒ b 5 0.
(3) is proved by straightforward calculation:
Let z15 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 *.
(4) can now be proved using the above results:
(z 1 (2z)) *5 0 * 5 0
but, (z 1 (2z)) *5 0 * 5 z * 1 (2z) *,
so z * 1 (2z) *5 0, and (2z) * 5 2z *.
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  ​5 (z *)21.
and (z21) * 5 __
z *
Conjugate zeros of polynomials
In Chapter 3, you used the following result without proof.
If c is a root of a polynomial equation with real coefficients, then c * is also a
root.
Theorem: If c is a root of a polynomial equation with real coefficients,
then c * is also a root of the equation.
We give the proof for n 5 3, but the method is general.
P(x) 5 ax 3 1 bx 2 1 dx 1 e
Since c is a root of P(x) 5 0, we have
ac 3 1 bc 2 1 dc 1 e 5 0
⇒ (ac 3 1 bc 2 1 dc 1 e) *5 0 Since 0* 5 0.
⇒ (ac ) * 1 (bc ) * 1 (dc) * 1 e * 5 0
Sum of conjugates theorem.
⇒ a(c *) 1 b(c *) 1 d(c *) 1 e 5 0
Result of product conjugate.
3
2
3
2
⇒ (c *) is a root of P(x) 5 0.
Example 8
1 1 2i is a zero of the polynomial P(x) 5 x 3 2 5x 2 1 11x 2 15. Find all
other zeros.
Solution
Since the polynomial has real coefficients, then 1 2 2i is also a zero. 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.
Now, P(x) 5 (x2 2 2x 1 5)(x 2 c). c can either be found by division or by
factoring by trial and error. In either case, c 5 3.
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.
Find all other zeros.
1
Not included in present IB syllabus.
437
10
Complex Numbers
Solution
Since the polynomial does not have real coefficients, then 1 2 2i is not
necessarily also a zero. 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). The
second factor can be factored into (x 1 1)(x 2 2 1 3i) giving us the other
two zeros as 21 and 2 2 3i.
Note: x2 1 (21 1 3i)x 2 2 1 3i 5 0 can be solved using the quadratic
formula.
_____________________
_______
1 2 3i 6 √
​  (21 1 3i)2 2 4(22
  
1 3i) ​
​ ______________________________
2b 6 √
​  b2 2 4ac  
     
​5 ​ 
 ​
   
  
x 5 ______________
​ 
2a
2
________________
1 2 3i 6 √
​  28 2 6i 1 8  
2 12i ​
_________________________
_____
1 2 3i 6 √
​  218i  
​
 ​
 ​
   
   5 ​ ______________
   
5 ​ 
2
2
_____
√
​we
​  218i  
let (a 1 bi)2 5 218i ⇒ a2 2 b2 1 2abi 5 218i, then
To find
equating the real parts and imaginary parts to each other: a2 2 b2 5 0
_____
and 2ab 5 218 will yield √
​  218i  
​5 63 7 3i, and hence
_____
1 2 3i 6 (63 7 3i)
1 2 3i 6 √
​  218i  
​ ________________
 ​
2 ​ 
 ​
x 5 ​ ______________
   
    
2
2
which will yield x 5 21 or x 5 2 2 3i.
Exercise 10.1
Express each of the following numbers in the form a 1 bi.
___
___
  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.
  7 (23 1 4i ) 1 (2 2 5i )
  8 (23 1 4i ) 2 (2 2 5i )
  9 (23 1 4i )(2 2 5i )
10 3i 2 (2 2 4i )
11 (2 2 7i )(3 1 4i )
12 (1 1 i )(2 2 3i )
3 1 2i 
13​ ______
 ​ 
2 1 5i
2 2 i  ​ 
14​ ______
3 1 2i
( 
( 
) ( 
) ( 
)
)
15​ __
​ 2 ​ 2 __
​ 1 ​ 1 __
​ 1 ​  i  ​ 1 ​ __
​ 1 ​  i  ​ 3 2
3 2
​ 2 ​ 2 __
​ 1 ​ 1 __
17​ __
​ 1 ​  i   ​ 4 ​ __
​ 1 ​  i  ​ 3 2
3 2
1 ​  (3 2 7i )
19​ __
i
13   ​ 
21​ _______
5 2 12i
438
( 
)( 
)
16​ __
​ 2  ​2 __
​ 2  ​1 __
​ 1 ​  i   ​​ __
​ 1 ​  i  ​ 3 2 3 2
18 (2 1 i )(3 2 2i )
20 (2 1 5i ) 2 (22 2 5i )
12i   ​ 
22​ ______
3 1 4i
( 
)
23 3i​ 3 2 __
​ 2 ​ i  ​ 3
39 2 52i 
 ​ 
25​ _______
24 1 10i
24 (3 1 5i )(6 2 10i )
26 (7 2 4i )21
3   ​ 1 ______
28​ ______
​  2   ​ 
3 2 4i 6 1 8i
_____
 
52√
​  2144 ​
____ ​ 
30​ __________
31√
​  216 ​ 
27 (5 2 12i )21
(7 1 8i )(2 2 5i )
  
 ​ 
29​ _____________
5 2 12i
31 Let z 5 a 1 bi. Find a and b if (2 1 3i )z 5 7 1 i.
32 (2 1 yi )(x 1 i ) 5 1 13i, where x and y are real numbers. Solve for x and y.
__
33 a) Evaluate (1 1 i​ √ 3 ​  )3.
__
b) Prove that (1 1 i​ √3 ​ )6n 5 82n, where n  Z1.
__
c) Hence, find (1 1 i​ √ 3 ​ )48.
__
__
34 a) Evaluate (2​ √ 2 ​ 1 i​ √2 ​ )2.
__
__
b) Prove that (2​ √2 ​ 1 i​ √2 ​ )4k 5 (216)k, where k  Z1.
__
__
c) Hence, find (2​ √2 ​ 1 i​ √ 2 ​ )46.
35 If z is a complex
number such that |z 1 4i | 5 2|z 1 i |, find the value of |z |.
______
( |z| 5 √
​  ​x2​ ​+ y​ 2​ ​ ​ 
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.
38 Find the complex number z and write it in the form a 1 bi if i(z 1 1) 5 3z 2 2.
39 Find the complex number z and write it in the form a 1 bi if ______
​  2 2 i  ​ 
√
​  z ​ 5 2 2 3i.
1 1 2i
_
40 Find the values of the two real numbers x and y such that (x 1 iy)2 5 3 2 4i.
41 a) Find the values of the two real numbers x and y such that (x 1 iy)2 5 2 8 1 6i.
b) Hence, solve the following equation
z2 1 (1 2 i )z 1 2 2 2i 5 0.
42 If z  C, find all solutions to the equation z3 2 27i 5 0.
43 Given that z 5 ​ _12 ​1 2i is a zero of the polynomial f (x) 5 4x3 2 16x2 1 29x 2 51,
find the other zeros.
44 Find a polynomial function
with integer coefficients and lowest possible degree
__
that has ​ _12 ​, 21 and 3 1 i​ √ 2 ​ as zeros.
45 Find a polynomial function __
with integer coefficients and lowest possible degree
that has 22, 22 and 1 1 i​ √3 ​ 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.
__
47 Given that z 5 1 2 i​ √3 ​ is a zero of the polynomial f (x) 5 3x3 2 4x2 1 8x 1 8,
find the other zeros.
z  ​  5 a 1 bi, show that |a 1 bi | 5 1.
48 Let z  C. If ​ __
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.
50 Solve the system of equations.
51 Solve the system of equations.
iz1 1 2z2 5 3 2 i
2z1 1 (2 1 i )z2 5 7 1 2i
10.2
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. Every ordered pair
(x, y) determines a unique complex number x 1 yi, and vice versa. This
correspondence is embodied in the geometric representation of complex
numbers. Looking at complex numbers as points in the plane equipped
with additional structure changes the plane into what we call complex
plane, or Gauss plane, or Argand plane (diagram). The complex plane has
two axes, the horizontal axis is called the real axis, and the vertical axis is
the imaginary axis. Every complex number z 5 x 1 yi is represented by a
point (x, y) in the plane. The real part is measured along the real axis and
the imaginary part along the imaginary axis.
imaginary axis
5i
3 4i
4i
3i
5 2i
2i
i
5
4
3
2
1
0
i
1
2
3
4
5 real axis
2i
3i
4i
The diagram above illustrates how the two complex numbers 3 1 4i and
25 1 2i are plotted in the complex plane.
z x 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.e. 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).
Since we are representing the complex numbers by vectors, this results in
some analogies between the two sets. So, adding two complex numbers or
subtracting them, or multiplying by a scalar, are similar in both sets.
Example 10
Consider the complex numbers z1 5 3 1 4i and z2 5 25 1 2i.
Find z1 1 z2 and z1 2 z2.
imaginary axis
6i
z1 z2 2 6i
4i
z1 z2 8 2i
2i
z1 3 4i
z2 5 2i
5
4
3
2
1
0
1
2
3 real axis
2i
Note here that the vector representing the sum, 22 1 6i, is the diagonal
of the parallelogram with sides representing 3 1 4i and 25 1 2i, while
the vector representing the difference is the second diagonal of the
parallelogram.
The length, norm, of a vector also has a parallel in complex numbers.
You recall that for a vector v 5 (x, y) the length of the vector is
______
|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 ​. 
Also of interest is the following result.
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 |.
441
10
Complex Numbers
Example 11
Calculate the moduli of the following complex numbers
a) z1 5 5 2 6i
b) z2 5 12 1 5i
Solution
______
___
a) |z1| 5 |5 2 6i | 5 √
​  52 1 62 ​ 
5√
​  61 ​ 
_______
____
b) |z2| 5 |12 1 5i | 5 √
​  122 1 52 ​ 
5√
​  169 ​ = 13
Example 12
Graph each set of complex numbers.
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. So, the set is a circle
with radius 3 and centre (0, 0)
as shown.
A
|z| 3
3
3
O
B
b) B is the set of complex numbers
whose distance from the origin is
less than or equal to 3. So, the set
is a disk of radius 3 and centre at
the origin.
|z| 3
3
O
3
Another important property is the following result:
|z1z2| 5 |z1| |z2|
Proof:
_________________________
​  (x1x2 2 y1y2)2   
1 (x1y2 1 x2y1)2 ​
|z1z2| 5 |(x1x2 2 y1y2) 1 (x1y2 1 x2y1)i| 5 √
_________________________________________________
2
2
2
5√
​  (x1x2)2 2 2x1x2y     
1y2 1 (y1y2) 1 (x1y2) 1 2x1y2x2y1 1 (x2y1)  ​
____________________________
2
5√
​  (x1x2)2 1 (y1y2)   
1 (x1y2)2 1 (x2y1)2 ​
But,
________
________
_________________
|z1| |z2| 5 √
​  x12 1 y12 ​ 
•√
​  x22 1 y22 ​ 
5√
​  (x12 1 y12)(x22  
1 y22) ​
____________________________
2
5√
​  (x1x2)2 1 (y1y2)   
1 (x1y2)2 1 (x2y1)2 ​
And so the result follows.
442
Example 13
Evaluate |(3 1 4i)(5 1 12i)|.
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 1  
562 ​5 √
​  4255 ​ 5 65
Trigonometric/polar form of a complex number
imaginary axis
r |z|
Imaginary part y
z x 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). Hence, using our knowledge of vectors, we
can introduce a new form for representing complex numbers – the
trigonometric form (also known as polar form).
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.
______
y
​  x2 1 y 2 ​ 
; and tan u 5 __
​ x ​.
Clearly x 5 r cos u and y 5 r sin u ; r 5 √
Therefore, z 5 x 1 yi 5 r cos u 1 (r sin u)i 5 r(cos u 1 i sin u).
The angle u is called the argument of the complex number, arg(z).
Arg(z) is not unique. However, all values differ by a multiple of 2p.
Note:
The trigonometric form is called ‘modulus-argument’ by the IB. Please
keep that in mind. 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. (cis u stands for cos u 1 i sin u.)
443
10
Complex Numbers
Example 14
Write the following numbers in trigonometric form.
__
a) z 5 1 1 i
b) z 5 √
​  3 ​ 2 i
c) z 5 25i
d) z 5 17
Solution
______
__
a) r 5 √
​  12 1 12 ​ 
5√
​  2 ​ ; tan u 5 __
​ 1 ​ 5 1.
1
Hence, by observing the real and
imaginary parts being positive,
we can conclude that the argument
must be u 5 __
​ p ​ .
4
__
__
p ​ 1 i sin ​ __
p ​   ​5 √
p ​ 
​  2 ​  cis ​ __
z5√
​  2 ​​ ( cos ​ __
4
4)
4
y
θ 11π
6
0
i
y
i
0
z1i
θ π4
x
____________
3
x
z 3i
__
__
2
__ ​ 
b) r 5 √
​  (​ √3 ​ )2 1 (21)
   ​5 √
​  4 ​ 5 2; tan u 5 ___
​ 21
. The real part is positive,
√
​  3 ​ 
the imaginary part is negative, and the point is therefore in the fourth
quadrant, so u 5 ____
 ​ 
. 
​ 11p
6
11p
11p
11p
 ​ 
 ​ 
 ​  
 1 i sin ​ ____
  ​5 2 cis ​ ____
 
 z 5 2​ cos ​ ____
8
6
6
p ​ .
We can also use u 5 2 ​ __
6
( 
)
c) r 5 5 and u 5 ___
​ 3p ​  since it is on the negative side of the imaginary axis.
2
y
3p ​    ​
3
p
 z 5 5​ cos ​ ___ ​  1 i sin ​ ___
2
2
p ​ .
We can also use u 5 2 ​ __
2
θ 3π
2
d) r 5 17 and u 5 0
0
( 
)
x
 z 5 17 (cos 0 1 i sin 0)
z 5 i
Example 15
Convert each complex number into its rectangular form.
4p ​  
a) z 5 3 cos 150° 1 3i sin 150°
b) z 5 12 cis ​ ___
3
p ​ 1 i sin ​ __
p ​   ​
c) z 5 6(cos 50° 1 i sin 50°)
d) z 5 15​( cos ​ __
2
2)
Solution
( 
__
)
(  )
__
2 ​ √3 ​ 
23​ √3 ​  __
3i
1
a) z 5 3​ _____
​   ​  
  ​1 3i​ ​ __  ​  ​5 _____
​   ​ 
 1 ​   ​ 
2
2
2
2
444
__
√
​  3 ​ 
4p ​  5 12  ​ ___
4p ​  1 12i sin ​ ___
21 ​ 1 12i  2 ​ ___
 ​ 5 26 2 ___
​ 6i__  
b) z 5 12 cos ​ ___
3
3
2
2
√
​  3 ​ 
 ​c) z 5 6 cos 50° 1 6i sin 50° 5 6  0.643 1 6i  0.766 5 3.857 1 4.596i
d) z 5 15(0 1 i ) 5 15i
Multiplication
The trigonometric form of the complex number offers a very interesting
and efficient method for multiplying complex numbers.
The analogy between complex
numbers and vectors stops at
multiplication. As you recall,
multiplication of vectors is
not ‘well defined’ in the sense
that there are two products
– the scalar product which is
a scalar, not a vector, and the
vector product (discussed
later) which is a vector but
is not in the plane! Complex
number products are complex
numbers!
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. Then
z1z25 (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)].
Now, using the addition formulae for sine and cosine, we have
z1z2 5 r1r2[(cos(u1 1 u2)) 1 i (sin(u1 1 u2))]
This formula says: To multiply two complex numbers written in
trigonometric form, we multiply the moduli and add the arguments.
Example 16
__
__
Let z1 5 2 1 2i ​ √3 ​ and z2 5 21 2 i ​ √3 ​. 
a) Evaluate z1z2 by using their standard forms (rectangular or Cartesian).
b) Evaluate z1z2 by using their trigonometric forms and verify that the two
results are the same.
Solution
__
__
__
__
__
a) z1z2 5 (2 1 2i ​ √3 ​ )(21 2 i √
​  3 ​)  5 (22 1 6) 1 (22​ √3 ​ 2 2​ √3 ​ )i 5 4 2 4i √
​  3 ​ 
b) Converting both to trigonometric form, we get
4p ​ ,  then
p ​ and z 5 2 cis ​ ___
z1 5 4 cis ​ __
2
3
3
(  ( 
))
(  )
( 
)
5p ​    ​
5p ​  1 i sin ​ ___
4p ​    ​  ​5 8 cis​ ___
z1z2 5 4  2​ cis​ __
​ p ​ 1 ​ ___
​ 5p ​    ​5 8​ cos ​ ___
3
3
3
3
3
(  (  ))
__
2​ √3 ​ 
_____
__
5 8​ __
​ 1 ​ 1 i​ ​   ​  
  ​  ​5 4 2 4i ​ √ 3 ​. 
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
4p ​ .  Alternatively, you can see it as multiplying z by z which results in
of ​ ___
2
1
3
a complex number whose magnitude is 4 times that of z2 and is rotated by
an angle of __
​ p ​ .
3
445
10
Complex Numbers
Example 17
__
Let z1 5 22 1 2i and z2 5 3​ √3 ​ 2 3i.
Convert to trigonometric form and multiply.
Solution
__
3p ​  and z 5 6 cis ​ ____
11p
 ​ 
,  then
z15 2​ √2 ​  cis ​ ___
2
4
6
(  ( 
(  )
))
7p ​   ​
7p ​ 1 i sin ​ ___
5 12​ √2 ​ cis​( ___
​ 7p ​  )​5 12​ √2 ​ ​( cos ​ ___
12
12
12 )
( 
)
__
__
__
11p
z1z25 12​ √2 ​ ​  cis​ ___
 ​  
​ 3p ​  1 ​ ____
  ​  ​5 12​ √ 2 ​  cis​ ____
​ 31p ​  
  ​5 12​ √ 2 ​  cis​ ___
​ 7p ​ 1 2p  ​
4
12
12
6
__
__
 
 
Note: You can simplify this answer further to get an exact rectangular form.
( 
(  ( 
)
)
)
__
7p ​   ​5 12​ √__
7p ​ 1 i sin ​ ___
3p 1  ​ 
3p 1  ​ 
4p 
4p 
2 ​​  cos​ ​ ________
 ​1 i sin ​ ________
 ​
z1z25 12​ √2 ​​  cos ​ ___
12
12
12
12
( 
__
p ​   ​1 i sin​  __
5 12​ √2 ​​  cos​( __
​ p ​ 1 ​ __
( ​ p4 ​ 1 ​ __p3 ​  )​  ​
4
3)
)
( (  2 2 2 2 ) (  2 2 2 2 ))
​  2 ​1 ​  6 ​
​  2 ​2 ​  6 ​ ________
 ​ 
 ​  
5 12​  2 ​​( ________
​ 
 1 i ​ 
 
)​5 (6 2 6​  3 ​) 1 i (6 1 6​  3 ​)
4
4
5
__
12​ √2 ​​ 
__
√
√
​  3 ​ 
​  2 ​  ___
___
__
__
__
√
​  2 ​ 
___
__
__
√
√
√
​  3 ​ 
​  2 ​  ___
​  2 ​  __
1 ​ 1 ___
​   ​  ​ 
​ ___
​   ​  ​ 
  1 ​ 2 ​   ​  ​ 
   ​   ​1 i ​ ​   ​  ​ 
  __
   ​   ​  ​
__
__ √  
√  
__
√  
__
√  
__
√  
__
√  
__
√  
Note: By comparing the Cartesian form of the product to the polar form,
__
__
__
__
__
__ √
√
​ 
​ 
2 ​
 2 √
​ 
6 ​
 
2 ​
 2 √
​ 
6 ​ 
7
7
p
p
___
___
________
________
 ​ 
 ​  
i.e. 12​ √2 ​​  cos ​   ​ 1 i sin ​   ​   ​and 12​ √2 ​ ​  ​ 
 1 i ​ 
 ​, we can
4
4
12
12
__
__
__
__
√
​  2 ​ 2 √
​  2 ​ 2 √
​  6 ​ 
​  6 ​ 
7p √
7p ​ 5 ​ ________
 ​ 
 ​ 
.
 and sin ​ ___ ​ 5 ​ ________
 
conclude that cos ​ ___
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.
You may have noticed that the conjugate of a
complex number z 5 r (cos u 1 i sin u) is
z* 5 r (cos u 2 i sin u) 5 r (cos(2u) 1 i sin(2u)).
imaginary axis
z x yi
Also, z  z*5 r (cos u 1 i sin u)  r (cos u 2 i sin u)
5 r2(cos2 u 1 sin2 u)
5 r2.
Graphically, a complex number and its conjugate
are reflections of each other in the real axis. See the
figure opposite.
θ
real axis
θ
z x yi
446
Division of complex numbers
A similar approach gives us the rules for division of complex numbers.
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. Then
cos u2 2 i sin u2
r1(cos u1 1 i sin u1) _____________
z1 ________________
   ​  ​ 
   ​
​ __
z2  ​5 ​ r2(cos u2 1 i sin u2) cos u2 2 i sin u2
( 
)
( 
)
r (cos u1 cos u2 1 sin u1 sin u2) 1 i(sin u1 cos u2 2 sin u2 cos u1)
​ 
     
   
 ​  ​
5 __
​ r1  ​​ ________________________________________________
2
cos2 u2 1 sin2 u2)
r (cos u1 cos u2 1 sin u1 sin u2) 1 i(sin u1 cos u2 2 sin u2 cos u1)
​ 
     
 
   ​  ​.
5 __
​ r1  ​​ ________________________________________________
1
2
Now, using the subtraction formulas for sine and cosine, we have
z1 __
r1
​ __
z2 ​ 5 ​ r2 ​ [(cos(u1 2 u2)) 1 i(sin(u1 2 u2))]
This formula says: To divide two complex numbers written in trigonometric
form, we divide the moduli and subtract the arguments.
In particular, if we take z1 5 1 and z2 5 z (i.e. u1 5 0 and u2 5 u), we will
have the following result.
1
1 ​5 __
1
__
If z 5 r(cos u 1 i sin u) then ​ __
z ​ r ​ (cos(2u) 1 i sin(2u)) 5 ​ r ​ (cos(u) 2 i sin (u))
Example 18
__
Let z1 5 1 1 i and z2 5 √
​  3 ​ 2 i.
a) Convert into trigonometric form.
1
b) Evaluate ​ __
z2  ​. 
z1
c) Evaluate ​ __
z2 ​ .
5p ​ .
5p ​ and cos ​ ___
d) Use the results above to find the exact values of sin ​ ___
12
12
Solution
__
2p
11p
p ​ ; z 5 2 cis ​ ____
 ​ 
 ​  
a) z1 1 √
​  2 ​  cis ​ __
 5 2 cis ​ ____
4 2
6
6
2p
p
1
1 __
1 __
____
__
b)​ __
z2  ​ 5 ​ 2 ​  cis​( 2 ​  6 ​   )​5 ​ 2 ​  cis ​ 6 ​ 
z1
1
__
c)​ __
z2 ​ can be found by either multiplying z1 by ​ z2  ​,  or by using division as
shown above.
( 
)
__
__
(  )
__
z1
√
√
​  2 ​  __
​  2 ​  ___
p ​   ​  ​ __
p ​   ​5 ___
p ​   ​5 ___
1
__
​  2 ​  cis ​ __
​ 1 ​  cis ​ __
​   ​ cis​
  ( ​ p ​ 1 ​ __
​   ​ cis​
  ​ 5p ​   ​, or
​ __
z2 ​ 5 z1  ​ z2  ​ 5 (​  √
4) 2
4
2
2
12
6
6)
__
p
__
__
__
√
​  2 ​  cis ​   ​  √
z1 _______
√
​  2 ​  ___
​  2 ​  __
2p
4 
 ​   )​5 ___
​ __
  ( ​ p ​ 2 ​ ____
​   ​ cis​
  ​ 5p ​   ​
 ​ 
5 ___
​   ​ cis​
2p
z2 ​ 5 ​ 2 cis ​ ____
4
2
2
12
6
 ​  
6
(  )
447
10
Complex Numbers
__
__
__
√
z
​  3 ​ 2 1 1 (​ √3 ​ 1 1)i
√
​  __
3 ​ 1 i _________________
 ​
d)​ __1 ​  5 ______
​  1__1 i   
 
   
​  ​ ______
 ​5 ​ 
z2
4
√
​  3 ​ 2 i √
​  3 ​ 1 i
Comparing this to part c).
__
__
__
__
__
√
√
√
√
​  3 ​ 2 1 ___
​  3 ​ 2 1 ___
​  6 ​ 2 √
​  2 ​ 
​  2 ​  ___
5p ​ 5 ​ ______
2
 ​ 
 ​  ​ 
 ​ 
 5 ​   ​ cos ​ 
  5p ​ ⇒ cos ​ ___
 
  __   ​ 5 ________
. 
​ ______
​ 
4
2
12
__
√
​  3 ​ 1 1
______
Also, ​ 
4
12
__
√
​  2 ​ 
___
5p ​ ⇒ sin ​ ___
5p ​ 5 ​ 
 ​ 
 5 ​   ​ sin ​ 
  ___
4
2
12
4
√
​  2 ​ 
__
__
√
​  3 ​ 1 1
______
12
__
√
​  6 ​ 1 √
​  2 ​ 
2__   ​ 5 ________
 ​  ​ 
 ​ 
 
  ___
. 
​ 
4
4
√
​  2 ​ 
Exercise 10.2
In questions 1–14, write the complex number in polar form with argument u, such
that 0 < u , 2p.
__
  1 2 1 2i
  2​ √3 ​ 1 i
  3 2 2 2i
  4​ √6 ​ 2 i √
​  2 ​ 
__
__
__
  5 2 2 2i √
​  3 ​ 
  6 23 1 3i
  7 4i
  8 23​ √3 ​ 2 3i
  9 i 1 1
10 215
11 (4 1 3i )21
12 i(3 1 3i )
13 p
14 ei
__
z
​ z1  ​.
In questions 15–24, find z1z2 and __
2
p ​ 1 i sin ​ __
p ​ , z 5 cos ​ __
p ​ 1 i sin ​ __
p ​ 
15 z1 5 cos ​ __
2
2
2
3
3
5p ​  1 i sin ​ ___
5p ​ ,  z 5 cos ​ ___
7p ​  1 i sin ​ ___
7p ​  
16 z1 5 cos ​ ___
2
6
6
6
6
p ​ 1 i sin ​ __
p ​ , z 5 cos ​ ___
2p ​  1 i sin ​ ___
2p ​  
17 z1 5 cos ​ __
6
6 2
3
3
13p ​ 
5p ​ 1 i sin ​ ___
13p, z 5 cos ​ ___
5p ​ 
 1 i sin ​ ____ ​ 
18 z1 5 cos ​ ____
12
12 2
12
12
3p ​  1 i sin ​ ___
3p ​    ​, z 5 __
4p ​  1 i sin ​ ___
4p ​    ​
​ 2 ​ ​  cos ​ ___
19 z1 5 3​ cos ​ ___
2
4
4
3
3
3
__
p
p
p
5
5
5
5p ​    ​
20 z1 5 3​ √ 2 ​ ​  cos ​ ___ ​  1 i sin ​ ___ ​    ​, z2 5 2​ cos ​ ___ ​  1 i sin ​ ___
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 ​  (cos 225° 1 i sin 225°), z2 5 ___
​   ​  (cos 330° 1 i sin 330°)
23 z1 5 __
8
2
__
24 z1 5 3​ √ 2 ​  (cos 315° 1 i sin 315°), z2 5 2(cos 300° 1 i sin 300°)
​ z1  ​ , __
​ 1  ​, 
In questions 25–30, write z1 and z2 in polar form, and then find the reciprocals __
z1
1 z2
__
the product z1z2, and the quotient ​ z   ​(2p , u , p).
__
__
2
25 z1 5 √
​  3 ​ 1 i and z2 5 2 2 2i √
​  3 ​ 
__
__
__
26 z1 5 √
​  6 ​ 1 i √
​  2 ​ and z2 5 2​ √3 ​ 2 6i
__
27 z1 5 4​ √ 3 ​ 1 4i and z2 5 23 2 3i
__
__
__
28 z1 5 i √
​  3 ​ and z2 5 2​ √ 2 ​ 2 i √
​  6 ​ 
__
__
__
29 z1 5 √
​  5 ​ 1 i √
​  5 ​ and z2 5 2i √
​  2 ​ 
448
__
__
30 z1 5 1 1 i √
​  3 ​ and z2 5 2​ √3 ​ 
31 Consider the complex number z where |z 2 i | 5 |z 1 2i |.
a) Show that Im(z) 5 2 ​ _12 ​.
b) Let z1 and z2 be the two possible values of z, such that |z | 5 1.
(i) Sketch a diagram to show the points which represent z1 and z2 in the
complex plane.
(ii) Find arg(z1) and arg(z2).
32 Use the Argand diagram to show that |z1 1 z2| < |z1| 1 |z2|.
( 
)
__
p ​   ​, express each of the following complex numbers in
2p ​  1 i sin 2​ __
33 If z 5 √
​  3 ​​  cos ​ ___
3
3
Cartesian form.
3
3 2 z2 ​ 
2z   ​ 
__    ​ 
a)​ ______
b)​ ______
c)​ ______
2
√
31z
3 1 z2
​  3 ​ 1 z
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 ).
35 If the numbers in question 34 represent the vertices of a triangle in the Argand
diagram, find the area of that triangle.
36 Identify, in the complex plane, the set of points that correspond to the following
equations.
a) |z | 5 3
b) z* 5 2z
c) z 1 z* 5 8
d) |z 2 3| 5 2
e) |z 2 1| 1 |z 2 3| 5 2
37 Identify, in the complex plane, the set of points that correspond to the following
inequations.
a) |z | < 3
b) |z 2 3i | > 2
10.3
Powers and roots of complex
numbers
The formula established for the product of two complex numbers can be
applied to derive a special formula for the nth power of a complex number.
Let z5 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).
Similarly,
z 35 z  z 2 5 (r (cos u 1 i sin u))(r 2(cos 2u 1 i sin 2u))
5 r 3(cos(u 1 2u) 1 i sin(u 1 2u)) 5 r 3(cos 3u 1 i sin 3u).
In general, we obtain the following theorem, named after the French
mathematician A. De Moivre (1667–1754).
449
10
Complex Numbers
Note: As a matter of fact, de Moivre stated ‘his’ formula only implicitly. Its
standard form is due to Euler and was generalized by him to any real n.
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).
The theorem: To find the nth power of any complex number written in trigonometric
form, we take the nth power of the modulus and multiply the argument with n.
Proof
The proof of this theorem follows as an application of mathematical
induction.
Let P(n) be the statement z n 5 r n(cos nu 1 i sin nu).
Basis step:
To prove this formula the basis step must be P(1).
P (1): is true since
z 1 5 r 1(cos u 1 i sin u), which is given!
[If you are not convinced, you can try
P (2): z 2 5 r 2(cos 2u 1 i sin 2u), which we showed above.]
Inductive step:
Assume that P(k) is true, i.e.
z k 5 r k(cos ku 1 i sin ku). We need to show that P(k 1 1) is also true.
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).
Now,
z k 1 1 5 z k  z 5 (cos ku 1 i sin ku)(r (cos u 1 i sin u)) by assumption
5 r kr [(cos kucos u 2 sin ku sin u) 1 i (sin ku cos u 1 cos ku sin u)]
5 r k 1 1[cos(ku 1 u) 1 i sin(ku 1 u)] by addition formulae for sine and
cosine
k11
5 r (cos(k 1 1)u 1 i sin(k 1 1)u)
Therefore, by the principle of mathematical induction, since the
theorem is true for n 5 1, and whenever it is true for n 5 k, it was
proved true for n 5 k 1 1, then the theorem is true for positive
integers n.
Note: In fact the theorem is valid for all real numbers n. However, the proof
is beyond the scope of this course and this book and therefore we will
consider the theorem true for all real numbers without proof at the moment.
Example 19
Find (1 1 i)6.
Solution
We convert the number into polar form first.
450
__
p ​ 1 i sin ​ __
p ​   ​
(1 1 i) 5 √
​  2 ​ ​ ( cos ​ __
4
4)
Now we can apply De Moivre’s theorem.
[ 
( 
__
p ​ 1 i sin ​ __
p ​   ​  6​​ ​5 (​ √__
p ​   ​1 i sin​  6  ​ __
(1 1 i)6 5 ​​ √
​  2 ​ ​ ( cos ​ __
2 ​)  6​ cos​( 6  ​ __
( p4 ​  )​  ​
4
4)
4)
3p ​    ​5 8(2i) 5 28i
3p ​  1 i sin ​ ___
5 8​ cos ​ ___
2
2
]
( 
)
)
Imagine you wanted to use the binomial theorem to evaluate the power.
(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.
For example, when n 5 21, the theorem gives the following result.
z21 5 r21(cos(2u) 1 i sin(2u)) 5 __
​ 1r ​ (cos u 2 i sin u)
Also,
z2n 5 (z21)n 5 (r21(cos(2u) 1 i sin(2u)))n 5 r2n(cos(2nu) 1 i sin(2nu)).
If we take the case when r 5 1, then
zn 5 cos nu 1 i sin nu and z2n 5 cos(2nu) 1 i sin(2nu) 5 cos nu 2i sin nu
⇒ zn 1 z2n 5 2 cos nu and zn 2 z2n 5 2i sin nu.
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.
Example 20
Find cos3 u in terms of first powers of the cosine function.
Solution
Starting with
( ​​ z 1 __​ 1z ​ )​​ ​5 (2 cos u)3
3
and expanding the left-hand side, we get
​ 13  ​ 5 8 cos3 u⇒ z3 1 __
​ 3z ​1 __
​ 13  ​ 1 3​( z 1 __
​ 1z ​ )​5 8 cos3 u
z3 1 3z 1 __
z z ⇕
⇕
⇒ 2 cos 3u 1 3(2 cos u) 5 8 cos3 u
⇒ cos3 u 5 ​ _18 ​ (2 cos 3u 1 3(2 cos u))
5 _​ 14 ​ (cos 3u 1 3 cos u)
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.
nth roots of a complex number
De Moivre’s theorem is an essential tool for finding nth roots of complex
numbers.
An nth root of a given number z is a number w that satisfies the following
relation
wn 5 z.
For example, w 5 1 1 i is a 6th root of z 5 28i because, as you have seen
above,
(1 1 i )6 5 28i, or
__
__
w 5 2​ √ 3 ​ 1 i is a 10th root of 512 1 512i √
​  3 ​ .
__
__
This is also because w 10 5 (2​ √3 ​ 1 i )10 5 512 1 512i √
​  3 ​ .
How to find the nth roots:
To find them, we apply the definition of an nth root as mentioned above.
Let w 5 s (cos a 1 i sin a) be an nth root of z 5 r (cos u 1 i sin u). This
means that w n 5 z, i.e.
(s(cos a 1 i sin a))n 5 r (cos u 1 i sin u) ⇒
s n(cos na 1 i sin na) 5 r (cos u 1 i sin u)
However, two complex numbers are equal if their moduli are equal, that is,
_
_1
sn 5 r ⇔ s 5 n​√  r  ​5 ​r ​ ​n ​​.
Also,
cos na 5 cos u and sin na 5 sin u.
From your trigonometry chapters, you recall that both sine and cosine
functions are periodic of period 2p each; hence,
cos na 5 cos u
⇒ na 5 u 1 2kp, k 5 0, 1, 2, ...
{ sin n
a 5 sin u
452
This leads to
u ​1 ____
  p 
​5 __
​ n
​ 2knp
  
​;  k 5 0, 1, 2, 3, ..., n 2 1.
a 5 _______
​ u 1n2k
 
Notice that we stop the values of k at n 2 1. This is so because for values
larger than or equal to n, principal arguments for these roots will be
identical to those for k 5 0 till 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  ​ 1 ____
u  ​ 1 ____
​ n
  ​ 1 i sin​ __
​ n
  ​  ​
zk 5 ​√  r ​  ​ cos​ __
​ 2knp
 ​  
​ 2knp
 ​  
where k 5 1, 2, 3, …, n 2 1.
_
1
__
Note: Each of the n nth roots of z has the same modulus ​√  r ​ 5 ​r​ ​n  ​​. Thus all these roots lie
n
1
__
_
on a circle in the complex plane whose radius is ​√  r ​ 5 ​r​ ​n  ​​. Also, since the arguments of
2p ​  , then the roots are also equally spaced on this circle.
consecutive roots differ by ​ ___
n
n
Example 22
Find the cube roots of z 5 28 1 8i.
Solution
__
r 5 8​ √2 ​ and u 5 ___
​ 3p ​ ,  so the roots are
4
(  ( 
)
( 
))
3p ​  
___
______
​ 3p ​  
​ ___
__
3
u
u
2k
2kp
2k
2k
p
p
p
4
w 5 ​ r ​​ cos​( __
​   ​  
 ​  
​  n   
​  )​1 i sin​( __
​  n   
​  )​  ​5 ​√  (​ 8  √
​ n ​1 ____
​ n ​1 ____
​  2 ​  )​ ​ ​ cos​ ___
​   ​ 1 ____
  ​1 i sin​ ___
​  4 ​ 1 ​ ____
  ​  ​
3
3
3
3
_
( 
)
n
√
   
(  ( 
( 
)
))
6 __
2kp
2kp
5 2​(​ √  2 ​  )​​ cos​ __
 ​  
 ​  
​ p ​ 1 ​ ____
  ​1 i sin​ __
​ p ​ 1 ​ ____
  ​  ​; k 5 0, 1, 2
4
4
3
3
6 __
w1 5 2​(​ √  2 ​  )​​ cos​( __
​ p ​  )​1 i sin​( __
​ p ​  )​  ​
4
4
__
6
2p ​    ​1 i sin​ __
2p ​    ​  ​5 2​6√  __
w2 5 2​(​ √  2 ​  )​​ cos​ __
​ p ​ 1 ​ ___
​ p ​ 1 ​ ___
2 ​​  cos​ ____
​ 11p ​  
  ​1 i sin​ ____
​ 11p ​  
  ​  ​
4
4
3
3
12
12
6 __
4p ​    ​1 i sin​ __
4p ​    ​  ​5 2​6√  __
w3 5 2​(​ √  2 ​  )​​ cos​ __
​ p ​ 1 ​ ___
​ p ​ 1 ​ ___
2 ​​  cos​ ____
​ 19p ​  
  ​1 i sin​ ____
​ 19p ​  
  ​  ​
4
4
3
3
12
12
( 
)
(  ( 
(  ( 
)
)
( 
( 
(  (  )
(  (  )
))
))
(  ))
(  ))
y
w1
w2
2π
3
2π
3
0
π
4
2π
3
x
w3
Notice how__the arguments are distributed equally around a circle with
6
2p ​ . 
radius 2​(​ √  2 ​  )​. The difference between any two arguments is ​ ___
3
453
10
Complex Numbers
Notice that if you try to go beyond k 5 2, then you get back to w1.
(  ( 
( 
)
))
( 
6 __
6p ​    ​1 i sin​ __
6p ​    ​  ​5 2​6√  __2 ​​  cos​  __
w4 5 2​√  2 ​​  cos​ __
​ p ​ 1 ​ ___
​ p ​ 1 ​ ___
( ​ p4 ​ 1 2p )​1 i sin​( __​ p4 ​ 1 2p )​  ​
4
4
3
3
6 __
5 2​√  2 ​​  cos​( __
​ p ​  )​1 i sin​( __
​ p ​  )​  ​5 w1
4
4
( 
)
)
Also, if you raise any of the roots to the third power, you will eventually get
z; for example,
(  ))
(  )) ]
(  (  )
[  (  (  )
5 8​ √2 ​​( cos​( ____
 ​  
 ​  
​ 11p
  ​1 i sin​( ____
​ 11p
  ​  ​5 8​ √ 2 ​​( cos​( ___
​ 3p ​   )​1 i sin​( ___
​ 3p ​   )​ )​5 z
4 )
4 ))
4
4
3
__
6 __
​ 11p ​  
  ​1 i sin​ ____
​ 11p ​  
  ​  ​  ​​ ​5 8​ √ 2 ​​  cos​ ____
​ 33p ​  
  ​1 i sin​ ____
​ 33p ​  
  ​  ​
(w2)35 ​​ 2​√  2 ​​  cos​ ____
12
12
12
12
__
__
 
 
Example 23
Find the six sixth roots of z 5 264 and graph these roots in the complex
plane.
Solution
Here r 5 64 and u 5 p. So the roots are
(  ( 
)
))
( 
u ​1 ____
u ​1 ____
​ 2knp
  
​   ​1 i sin​ __
​ 2knp
  
​   ​  ​
​ n
​ n
w 5 s​ cos​ __
(  ( 
(  ( 
( 
)
))
__
6
2k ​ 
2kp
p 
 ​  
​ p ​ 1 ​ ____
  ​1 i sin​ __
​ p ​ 1 ​ ____
  ​  ​
5√
​  64 ​​  cos​ __
6
6
6
6
kp ​    ​1 i sin​ __
kp ​    ​  ​; k 5 0, 1, 2, 3, 4, 5
​ p ​ 1 ​ ___
​ p ​ 1 ​ ___
5 2​ cos​ __
3
3
6
6
( 
w3
w2
3
2
1
w4
1
( 
( 
0
1
2 w
5
π
6
1
2
w6
3
x
)
)
)
( 
))
(  ( 
(  ))
(  (  )
3p ​    ​1 i sin​  __
3p ​    ​  ​
w 5 2​( cos​( __
​ p ​ 1 ​ ___
(​ p6 ​ 1 ​ ___
3 )
3 ))
6
7p ​    ​1 i sin​  ___
5 2​( cos​( ​ ___
(​ 76p ​   )​ )​
6 )
4p ​    ​1 i sin​  __
4p ​    ​  ​
w 5 2​( cos​( __
​ p ​ 1 ​ ___
(​ p6 ​ 1 ​ ___
3 )
3 ))
6
​ 3p ​   )​1 i sin​( ___
​ 3p ​   )​ )​
5 2​( cos​( ___
2
2
5p ​    ​1 i sin​  __
5p ​    ​  ​
w 5 2​( cos​( __
​ p ​ 1 ​ ___
(​ p6 ​ 1 ​ ___
3 )
3 ))
6
11p
 ​  
 ​  
​ 11p
  ​1 i sin​( ​ ____
  ​  ​
5 2​( cos​( ____
6 )
6 ))
2p ​    ​1 i sin​ __
2p ​    ​  ​
w3 5 2​ cos​ __
​ p ​ 1 ​ ___
​ p ​ 1 ​ ___
3
3
6
6
5
5
p
p
___
___
5 2​ cos​ ​   ​    ​1 i sin​ ​   ​    ​  ​
6
6
4
5
6
454
)
p ​   ​1 i sin​  __
w2 5 2​ cos​( __
​ p ​ 1 ​ __
( ​ p6 ​ 1 ​ __p3 ​  )​  ​
3)
6
​ p ​  )​1 i sin​( __
​ p ​  )​  ​
5 2​ cos​( __
2
2
w1
π
3
))
w1 5 2​ cos​( __
​ p ​  )​1 i sin​( __
​ p ​  )​  ​
6
6
y
2
( 
)
nth roots of unity
The rules we established can be applied to finding the nth roots of 1
(unity). Since 1 is a real number, then in polar/trigonometric form it has a
modulus of 1 and an argument of 0. We can write it as
1 5 1(cos 0 1 i sin 0).
Now applying the rules above, 1 has n distinct nth roots given by
)
( 
))
(  ( 
0 ​1 ____
0 ​1 ____
​ 2knp
  
​  )​1 i sin​( __
​ 2knp
  
​  )​ )​
​ n
​ n
5 ​√  1 ​ (​  cos​( __
  
​  )​1 i sin​( ____
  
​  )​; k 5 0, 1, 2, ..., n 2 1
​ 2knp
​ 2knp
5 cos​( ____
_
n
u ​1 ____
u ​1 ____
​ 2knp
  
​   ​1 i sin​ __
​ 2knp
  
​   ​  ​
​ n
​ n
zk 5 ​√  r  ​ cos​ __
n
__
 
Or in degrees,
( 
)
( 
)
​ 360k
 ​  ​1 i sin​ ____
​ 360k
 ​  ​; k 5 0, 1, 2, ..., n 2 1
zk 5 cos​ ____
n   
n   
Example 24
Find
a) the square roots of unity
b) the cube roots of unity.
Solution
a) Here k 5 2, and therefore the two roots are
(  )
(  )
(  )
(  )
(  )
(  )
 ​  
 ​  
​ 360k
  ​1 i sin​ ____
​ 360k
  ​; k 5 0, 1
zk 5 cos​ ____
2
2
​ 0 ​   ​1 i sin​ __
​ 0 ​   ​5 1
z0 5 cos​ __
2
2
360
​   ​    ​1 i sin​ ___
​ 360 ​    ​5 cos 180 1 i sin 180 5 21
z1 5 cos​ ___
2
2
b) Here k 5 3, and the three roots are
(  )
(  )
(  )
(  )
(  )
(  )
(  )
(  )
 ​  
 ​  
​ 2kp
  ​1 i sin​ ____
​ 2kp
  ​; k 5 0, 1, 2, 3
zk 5 cos​ ____
3
3
​ 0  ​  ​1 i sin​ __
​ 0 ​   ​5 1
z0 5 cos​ __
3
3
__
√
​  3 ​ 
2
2
p
p
1
___
___
__
___
z1 5 cos​ ​   ​    ​1 i sin​ ​   ​    ​5 2 ​   ​ 1 i ​   ​ 
3
3
2
2__
√
​  3 ​ 
1 ​ 2 i ​ ___
 ​ 
​ 4p ​    ​1 i sin​ ___
​ 4p ​    ​5 2 ​ __
z2 5 cos​ ___
3
3
2
2
Euler’s formula
The material in this part depends on work that you will do in the Analysis
option. Otherwise, you will have to accept the result without proof.
455
10
Complex Numbers
In the options section on infinite series, we have the following results.
Taylor’s (Maclaurin’s) series expansion for sin x, cos x and e x are
3
5
7
3!
5!
7!
​ x  ​ 2 __
​ x  ​ 1 ... 5
sin x 5 x 2 __
​ x  ​ 1 __
∞
2n 1 1
​  x   ​ 
∑(21)n ________
(2n 1 1)!
0
∞
2
4
6
2n
​ x  ​ 2 __
​ x  ​ 1 ... 5 ∑(21)n _____
​  x   ​ 
cos x 5 1 2 __
​ x  ​ 1 __
2! 4! 6!
(2n)!
0
2
3
4
2!
3!
4!
​ x  ​ 1 __
​ x  ​ 1 ... 5
​ x  ​ 1 __
e 5 1 1 x 1 __
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 __
​ x  ​ 2 __
​ x  ​ 1 __
​ x  ​ 1 __
​ x  ​ 2 __
​ x  ​ 2 __
​ x  ​ 1 ...
2! 3! 4! 5! 6! 7!
and compare the result to e x expansion, we notice a stark similarity in
the terms, except for the ‘discrepancy’ in the signs! The signs in the sum
alternate in a way where pairs of terms alternate! This property is typical of
powers of i.
Look at i, i 2, i 3, i 4, i 5, i 6, i 7, i 8, ... 5 i, 21, 2i , 1, i , 21, 2i , 1, …
This suggests expanding e ix
2 2
3 3
4 4
5
6 6
i 5x  ​
 1 ​ ___
​ i x  ​ 1 ____
​ i x  ​
​ i x  ​ 1 …
e ix 5 1 1 ix 1 ___
​ i x  ​ 1 ___
 1 ____
2!
3!
4!
5!
6!
2 2
4 4
6 6
3 3
5 5
5 1 1 ___
​ i x  ​ 1 ____
​ i x  ​ 1 ix 1 ___
​ i x  ​  ! 1 ____
​ i x  ​ 1 ___
​ i x  ​ 1 …
4
2!
6!
3!
5!
( 
)
x  ​ 1 __
5 1 2 ​ __
​ x  ​ 2 ___
​ x   ​ 1 … 1 i (​  x 2 __
​ x  ​ 1 … )​
​ x  ​ 1 __
2! 4! 6!
3! 5!
2
4
6
2 3
4 5
5 1 2 __
​ x  ​ 1 __
​ x  ​ 2 __
​ x  ​ 1 … 1 i ​ x 1 ___
​ i x  ​ 1 …  ​
​ i x  ​ 1 ____
2! 4! 6!
3!
5!
2
4
6
3
5
5 cos x 1 i sin x
Since, for any complex number
z 5 x 1 iy 5 r (cos u 1 i sin u) and since e iu 5 cos u 1 i sin u, then
z 5 r (cos u 1 i sin u) 5 re iu.
This is known as Euler’s formula.
Example 25
Evaluate each of the following
a) e ip
p
__
b)​e i ​ ​ 2  ​​
Solution
a) e ip 5 cos p 1 i sin p 5 21
p
__
p ​ 1 i sin ​ __
p ​ 5 i
b)​e i ​ ​ 2  ​​5 cos ​ __
2
2
456
Example 26
Use Euler’s formula to prove DeMoivre’s theorem.
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​e i ​ ​ 6  ​​
b) z 5 7e 2i
Solution
__
3​ √ 3 ​ 
p ​ 5 ​ ____
 ​  and
a) Since |z | 5 3 and arg(z) 5 __
​ p ​ , Re(z) 5 3 cos ​ __
2
6
6
3
p
Im(z) 5 3 sin __
​   ​ 5 ​ __  ​.
6 2
b) Since |z | 5 7 and arg(z) 5 2, Re(z) 5 7 cos 2 and Im(z) 5 7 sin 2.
Example 28
Express z 5 5 1 5i in exponential form.
Solution
__
__ __p
|z | 5 5​ √2 ​ and tan u 5 __
​ 5 ​ 5 1 ⇒ u 5 __
​ p ​ , therefore z 5 5​ √ 2 ​  ​e i ​ ​ 4  ​​.
5
4
Example 29
Evaluate (5 1 5i)6 and express your answer in rectangular form.
Solution
  __
p
__
Let z 5 5 1 5i. From the example above, z 5 5​√  2 ​ ​ e i ​ ​ 4  ​​; hence,
z 6 5 (​​  5​√  2 ​ ​ e i ​ ​ 4  ​​ )​​ ​5 (5​ √2 ​ )6​e i ​ ​ 4  ​​ 3 6 5 125 000 ​ei ​ ​ 2  ​​5 2125 000i.
  __
p 6
__
__
p
__
3p
__
Alternatively,
( 
( 
)
__
6p ​    ​5 2125 000i.
6p ​  1 i sin ​ ___
p ​ 1 i sin ​ __
p ​   ​  6​​ ​5 (5​ √__
(5 1 5i)6 5 ​​ 5√
​  2 ​​(   cos ​ __
2 ​ )6​ cos ​ ___
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) _______
_____________________________
    ​5 ​ e  e ​ 
   
   
 
​5 e 5iu 5 cos 5u 1 i sin 5u
4iu
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.
Solution
__
___
__
__
_
​   ​
i 5 e​ i​​( ​ 2  ​1 2kp )​​⇒ ​i ​ ​3 ​​5 (​​  ​e i​​( ​ 2  ​1 2kp )​​ )​3​5 ​e i​​( ​ 6  ​1 ​  3   ​  )​​; k 5 0, 1, 2
p
1
_1
p
Therefore,
2kp
p
__
p
__
√
​  3 ​  __i
p ​ 1 i sin ​ __
p ​ 5 ​ ___
z0 5 e​ ​i(​  ​ 6  ​ )​​5 cos ​ __
 ​ 1 ​    ​ 
2
2
6
6
__
2p
5p
__
√
​  3 ​  __i
i ​(  __
​ p6  ​ 1 ​ __
 
 
​ 
​
5
5
p
p
i ​( ​  6  ​ )​
)
___
___
___
3
​5 ​e ​ ​5 cos ​   ​  1 i sin ​   ​  5 2 ​   ​ 1 ​    ​ 
z1 5 e​ ​
2
2
6
6
p
4p
3p
__
__
__
3p ​  5 2i
3p ​  1 i sin ​ ___
z2 5 ​e ​i ​(  ​ 6  ​ 1 ​  3  ​ )​​5 ​e i ​​( ​ 2  ​ )​​5 cos ​ ___
2
2
As you notice here, Euler’s formula provides us with a very powerful tool to
perform otherwise extremely laborious calculations.
Exercise 10.3
In questions 1–6, write the complex number in Cartesian form.
2p
__
  1 z 5 4​e2i ​ 
​ 3  ​ ​
  2 z 5 3e2pi
  3 z 5 3e0.5pi
  4 z 5 4 cis​ ___
​ 7p ​   ​(exact value)
12
11 ​ __p3  ​  i
  6 z 5 3​e​ ​
(  )
__
​ pi  ​ 
  5 z 5 13​e​3 ​
In questions 7–16, write each complex number in exponential form.
__
  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. Express in exact rectangular form
when possible.
__
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.
__
26 The square roots of 4 1 4i​ √ 3 ​ .
__
27 The cube roots of 4 1 4i​ √ 3 ​ .
28 The fourth roots of 21.
29 The sixth roots of i.
__
30 The fifth roots of 29 2 9i​ √ 2 ​ .
458
In questions 31–36, solve each equation.
31 z 5 2 32 5 0
32 z 8 1 i 5 0
__
33 z 3 1 4​ √3 ​ 2 4i 5 0
34 z 4 2 16 5 0
35 z 5 1 128 5 128i
36 z 6 2 64i 5 0
In questions 37–40, use De Moivre’s theorem to simplify each of the following
expressions.
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 ​​
__________________
40​n√  (cos(2nb) 1 i sin(2n
  
b)) ​
41 Use e iu to prove that cos(a 1 b) 5 cos a cos b 2 sin a sin b.
42 Use De Moivre’s theorem to show that cos 4a 5 8 cos4 a 2 8 cos2 a 1 1.
43 Use De Moivre’s theorem to show that cos 5a 5 16 cos5 a 2 20 cos3 a 1 5 cos a.
44 Use De Moivre’s theorem to show that cos4 a 5 _​ 18 ​ (cos 4a 1 4 cos 2a 1 3).
45 Let z 5 cos 2a 1 i sin 2a.
1 ​  5 2 cos 2a and that 2i sin 2a 5 z 2 __
a) Show that z 1 ​ __
​ 1z ​ .
z
b) Find an expression for cos 2na and sin 2na in terms of z.
46 Let the cubic roots of 1 be 1, v and v2. Simplify (1 1 3v)(1 1 3v2).
47 a) Show that the fourth roots of unity can be written as 1, b, b2, and b3.
b) Simplify (1 1 b)(1 1 b2 1 b3).
c) Show that b 1 b2 1 b3 5 21.
48 a) Show that the fifth roots of unity can be written as 1, a, a2, a3 and a4.
b) Simplify (1 1 a)(1 1 a4).
c) Show that 1 1 a 1 a2 1 a3 1 a4 5 0.
__
__
49 Show that (1 1 i √
​  3 ​ )n 1 (1 2 i √
​  3 ​ )n is real and find its value for n 5 18.
50 Given that z 5 (2a 1 3i )3, and a  1, find the values of a such that arg z 5 135°.
Practice questions
  1 Let z 5 x 1 yi. Find the values of x and y if (1 2 i )z 5 1 2 3i.
  2 Let x and y be real numbers, and v be one of the complex solutions of the equation
z 3 5 1. Evaluate:
a) 1 1 v 1 v2
b) (vx 1 v2y)(vy 1 v2x)
  3 a) Evaluate (1 1 i )2.
b) Prove, by mathematical induction, that (1 1 i )4n 5 (24)n, where n  N+.
c) Hence or otherwise, find (1 1 i )32.
459
10
Complex Numbers
__
__
√
 
i ​ √2 ​  
​  6 ​ 2
  4 Let z1 5 ​ ________
  z2 5 1 2 i.
 ​ and
2
p ​ <
p ​ .
a) Write z1 and z2 in the form r (cos u 1 i sin u), where r > 0 and 2 ​ __
  u < ​ __
2
2
z1
p  ​  1 i sin ​ ___
p  ​ .
___
 
 ​ 5
cos ​ 
b) Show that ​ __
z2
12
12
z1
 
 ​ i
n
the
form
a 1 bi, where a and b are to be determined exactly
c) Find the value of ​ __
z2
p  ​  and sin ​ ___
p  ​ .
in radical (surd) form. Hence or otherwise, find the exact values of cos ​ ___
12
12
( 
( 
)
p ​ 1
p ​   ​ and z 5 b​ cos ​ __
p ​ 1
p ​   ​.
  i sin ​ __
  i sin ​ __
  5 Let z1 5 a​ cos ​ __
2
4
4)
3
3
z1 3
__
Express ​​ ​ z  ​   ​​ ​ in the form z 5 x 1 yi.
2
(  )
  6 If z is a complex number and |z 1 16| 5 4|z 1 1|, find the value of |z |.
  7 Find the values of a and b, where a and b are real, given that (a 1 bi )(2 2 i ) 5 5 2 i.
  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.
10 a) Express z 5 2 1 as a product of two factors, one of which is linear.
b) Find the zeros of z 5 2 1, giving your answers in the form
r (cos u 1 i sin u ) where r . 0 and 2p , u < p.
4
c) Express z 1 z 3 1 z 2 1 z 1 1 as a product of two real quadratic factors.
11 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.
( 
) ( 
)
p ​ 2
p ​   2​​ ​​​ cos ​ __
p ​ 1
p ​   3​​ ​
  i sin ​ __
  i sin ​ __
​​ cos ​ __
4
4
3
3
_____________________________
12 Consider the complex number z 5 ​ 
 ​ .
   
   
4
p
p
___
___
​​ cos ​    ​  2 i sin ​    ​   ​​ ​
( 
24
24
)
a) (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, i.e. z 5 ​√  1 ​ .
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.
2   ​ + 1 – 4i.
13 The complex number z satisfies the equation ​ √ z ​ = ​ ____
1–i
Express z in the form x 1 i y where x, y  Z.
_
14 a) Prove, using mathematical induction, that for a positive integer n,
(cos u 1 i sin u)n 5 cos nu 1 i sin nu where i 2 5 21.
b) The complex number z is defined by z 5 cos u 1 i sin u.
1 ​  5 cos(2u) 1 i sin(2u).
(i) Show that ​ __
z
(ii) Deduce that z n 1 z2n 5 2 cos nu.
c) (i) Find the binomial expansion of (z 1 z 21)5.
1
  ​ (a cos 5u 1 b cos 3u 1 c cos u), where a, b and
(ii) Hence, show that cos5 u 5 ​ __
16
c are positive integers to be found.
460
15 Consider the equation 2(p 1 iq) 5 q 2 ip 2 2(1 2 i ), where p and q are both real
numbers. Find p and q.
16 Consider z 5 2 32 5 0.
( 
)
2p ​ 1
2p ​   ​ is one of the complex roots of this equation.
(i) Show that z1 5 2​ cos ​ ___
  i sin ​ ___
5
5
(ii) Find z 12, z 13, z 14 and z 15 giving your answer in the modulus argument form.
(iii) Plot the points that represent z1, z 12, z 13, z 14 and z 15 in the complex plane.
(iv) The point z n1 is mapped to z n 1 11 by a composition of two linear transformations,
where n 5 1, 2, 3, 4. Give a full geometric description of the two transformations.
17 A complex number z is such that |z | 5 |z 2 3i |.
3 ​ .
a) Show that the imaginary part of z is ​ __
2
b) Let z1 and z2 be the two possible values of z, such that |z | 5 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.
p ​ .
(ii) Show that arg(z1) 5 ​ __
6
(iii)Find arg(z2).
(  )
z k z2
​  1  ​ 
 ​ 5 p, find a value for k.
c) Given that arg​ ____
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.
19 Consider the complex number z 5 cos u 1 i sin u.
a) Using De Moivre’s theorem show that
1  ​ 5 2 cos nu.
z n 1 ​ __
z n
1 ​    4​​ ​ show that
b) By expanding ​​ z 1 ​ __
z
( 
)
cos4 u 5 ​ _18 ​ (cos 4u 1 4 cos 2u 1 3).
20 Consider the complex geometric series e iu 1 ​ _12 ​ e2iu 1 ​ _14 ​ e3iu 1 …
a) Find an expression for z, the common ratio of this series.
b) Show that |z | , 1.
c) Write down an expression for the sum to infinity of this series.
d) (i) Express your answer to part c) in terms of sin u and cos u.
(ii) Hence, show that
4 cos u 2 2  
cos u 1 ​ _12 ​ cos 2u 1 ​ _14 ​ cos 3u 1 … 5 ​ _________
​ .
5 2 4 cos u
21 Let P (z) 5 z 3 1 az 2 1 bz 1 c, where a, b and c  . Two of the roots of P (z ) 5 0 are
22 and (23 1 2i ). Find the value of a, of b and of c.
__
22 Given that |z | 5 2​ √ 5 ​ , find the complex number z that satisfies the equation
15  ​ 5 1 2 8i.
___
​ 25  ​ 2 ​ ___
*z*
z*
23 Solve the simultaneous system of equations giving your answers in x 1 i y form:
iz1 1 2z2 5 3
z1 1 (1 2 i )z2 5 4
461
10
Complex Numbers
24 a) Solve the equation x2 2 4x 1 8 5 0. Denote its two roots by z1 and z2 and
express them in exponential form with z1 in the first quadrant.
z 4
b) Find the value of ​ __21 ​  and write it in the form x 1 yi.
z 2
c) Show that z 14 5 z 24.
z1 __
z2
d) Find the value of ​ __
z2 ​  1 ​ z1 ​ .
e) For what values of n is z 1n real?
2p ​ 1
2p ​ is
  i sin ​ ___
  a root of the equation x7 2 1 5 0.
25 a) Show that z 5 cos ​ ___
7
7
b) Show that z 7 2 1 5 (z 2 1)(z 6 1 z 5 1 z 4 1 z 3 1 z 2 1 z 1 1) and deduce that
z 6 1 z 5 1 z 4 1 z 3 1 z 2 1 z 1 1 5 0.
2p ​ 1
4p ​ 1
6p ​ 5
c) Show that cos ​ ___
  cos ​ ___
  cos ​ ___
  2 ​ _12 ​.
7
7
7
Questions 1–23 © International Baccalaureate Organization
462