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BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
COMPLEX NUMBERS
Jerome Cardan is widely credited as being the first person to create the concept of complex
numbers, to solve previous unsolvable problems, in 1545.
He stated that we can think of ( 1) as having a solution i.
i  ( 1)
Examples:
1.
25  25  (1)  5 1  5i
2.
7  7  (1)  7  1  7i
A complex number has two parts to it; a real part and an imaginary part.
z  a  bi
We will define complex number as
where a  R( z) and b  I ( z) .
For example, z1  2  3i and z2  5  7i are both complex numbers.
It is worth noting that
i  ( 1) , i 2  1, i3  i, i 4  1, i5  i etc.
Also, we can show that if
z1  a  bi and z2  c  di and z1  z2 then a  c and b  d .
For z1  a  bi and z2  c  di where a, b, c, d  R we can observe the following results:
(i)
z1  z2   a  bi    c  di 
(ii)
 a  c  bi  di
z1z2   a  bi  c  di 
 ac  adi  bci  bdi 2
  a  c   b  d  i
 ac  bd  adi  bci
  ac  bd    ad  bc  i
which is a complex number.
which is a complex number.
(iii)
 z1 
2
  a  bi 
2
(iv)
where k  R
 ka  kbi
 a 2  2abi  b 2i 2
  ka    kb  i
which is a complex number.
 a 2  b 2  2abi

kz1  k  a  bi 

 a 2  b 2   2ab  i
which is a complex number.
Now try:
Page 90, Exercise 1 – questions 1, 2(f to i), 4f, 5, 6i, 7b, 8
PAGE 1 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Complex numbers can be used when solving quadratic equations.
Example:
3.
x2  4x  5  0
discriminant  b2  4ac
 4 2  4  1 5
 4
 roots are distinct and not real (i.e. complex).
2
x  b  b  4ac
2a
 4   4
2
 4  2i
2
 2  i or  2  i
4.
Find all three roots that satisfy x3  6 x 2  12 x  7  0 .
PAGE 2 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
The Complex Conjugate is used to rationalise denominators etc.
Let z  a  bi .
Examples:
Result (v):
Therefore, the conjugate of z is, z  a  bi .
5.
The conjugate of 3  i is 3  i .
6.
z  5  3i
 z  5  3i .
Therefore, for z  a  bi and z  a  bi (where a, b  R ) we get the following
result:
zz   a  bi  a  bi 
[a difference of two sqaures]
  a    bi 
2
2
 a 2  b 2i 2
 a 2  b2
[which is a real number]
Example:
7.
To rationalise the denominator of a complex fraction:
3  2i  3  2i  1  2i
1  2i 1  2i 1  2i
 3  2i 1  2i 

1  2i 1  2i 
2
 3  8i  4i2
12   2i 
 3  8i  4
1 4
 1  8i
5
1 8i
5 5
PAGE 3 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
We can use the above results to calculate the square root of a complex number:
Example:
8.
We know that 3  4i  a  bi
because  complex number    complex number 
2
Hence,
3  4i  a  bi
[see result (iii) above]
where a, b  R .
Square both sides to give
3  4i   a  bi 
2


 3  4i  a 2  b 2  2abi
By equating real and imaginary coefficients:
 a 2  b2   3
2ab  4
equation 1
b 2
a

equation 2
Substitute equation 2 into equation 1 to get,
 
a2   2
a
2
3
a 2  42  3
a
4
a  4 3
a2 a2
a4  4  3
a2
a 4  4  3a 2
a 4  3a 2  4  0
 a 2  4  a 2  1  0
 a2  4
a  2
or
a 2  1
a   i invalid as a  R
When a  2, b  1 and
Therefore,
a  2, b  1
3  4i  2  i or 2  i .
Now try:
Page 91, Exercise 2 – questions 3e, 5a, 2(c, e), 3(a, c, f), 4b, 5b, 6b(iv, v), 4c, 5c, 6b(vi), 8
PAGE 4 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Solving Equations with Complex Roots
In example 3 above we saw that for x 2  4 x  5  0 we have roots that are distinct and not real (i.e.
complex).
2
x  b  b  4ac
2a
 4   4
2
 2  i or  2  i
Notice that when we use the quadratic formula to solve an equation we will always get two roots: in
the above example we obtain the roots z  2  i and z  2  i .
In fact, for any equation with a complex root the conjugate of that complex number will also
be a root.
These roots provide us with factors and the product of the two factors will also be a factor
(think of the prime factors of 20. 20  2  2  5 . The product of any two of these factors is also a
factor).
Example:
9.
Given that 4  i is a root of z 4  8z3  13z 2  32z  68  0 , find all other roots that satisfy
this equation.
[N.B. to check that 4  i is a root we could substitute it into the equation and observe that it
works out to be zero.]
  z  (4  i)  is a factor
z  4  i is a root
so z  4  i is a root
  z  (4  i)  is also a factor
and  z  (4  i)  z  (4  i)  is also a factor
i.e.
 z  (4  i)  z  (4  i)    ( z  4)  i  ( z  4)  i 
 ( z  4)2  i 2
 z 2  8 z  16  (1)
 z 2  8 z  17 is also a factor
We can use this quadratic factor to divide the polynomial
z2
4
z 2  8 z  17 z 4  8 z 3  13 z 2  32 z  68
z 4  8 z 3  17 z
 4 z 2  32 z  68
 4 z 2  32 z  68
0
PAGE 5 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS


z 4  8 z 3  13z 2  32 z  68  z 2  8 z  17 z 2  4




 z 2  8 z  17 z 2  4  0
10.
 z 2  8 z  17  0
or
z2  4  0
 z  4  i, z  4  i
or
z  2, z  2
Find all the roots of 2z 4  3z3  17 z 2  12z 10  0 given that ( z  1  3i ) is a factor of
2z 4  3z3  17 z 2  12z 10 .
Now try:
Page 108, Exercise 8 – questions 4, 5, 6b, 7b, 8
PAGE 6 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Argand Diagrams
A complex number can be represented geometrically by using an Argand diagram (displaying the
complex number as we would a vector).
The x-axis represents the real part of the complex number and the y-axis represents the imaginary
part.
This plane is called the complex plane and the complex number x + yi is represented by the point
P(x, y).
P
y (Imaginary)
r
y

O
x
x (Real)
P(x, y) represents the complex number x  yi .
OP is considered to be a vector that has been rotated off the x-axis, the length of OP is r.
The size of the rotation is called the argument of z and is often denoted by Arg z (for the diagram
above notice that Arg z is   2n where n  Z ).
We call the value of Arg z that lies in the range      the principal argument of z and denote
this as arg z (notice the lower case is used here).
r  z , the modulus of z and so,
r  x2  y 2
We can see also that
and
and therefore
  tan 1 
y

x
 
where      
x  r cos
y  r sin 
z  r  cos  i sin  
z  x  yi but written in polar form z  r  cos  i sin   .
PAGE 7 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Examples:
11.
y
O
P
r  32  22  13 = 3  61 to 3 sig. fig.
2
Arg z  tan 1 2  2n
3
 

x
3
(3, 2) is in the first quadrant so arg z = 0 187 radians (to 3 s.f.)
Sketch the Argand diagram for z = -4 – 3i and find modulus and argument of z.
12.
y
-4

-3
P
r  (4)2  (3) 2  5
O
x
 
Arg z  tan 1 3  2n (remember,      ).
4
(-4, -3) is in the third quadrant so
arg z = (0  205   )  0  795 radians (to 3 s.f.)
or arg z = (36  9  180)  143 (to 3 s.f.)
Now try:
Page 94, Exercise 3 – questions 3(f, h), 4, 6(a, e), 7(b, c, i, h), 8
PAGE 8 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
The Polar Form of Complex Numbers
We have already seen that complex numbers can be written as
z1  r  cos  i sin  
z2  s  cos   i sin  
and
Using the polar form of complex numbers we get the following results:
(vi)
z1 z2  r  cos  i sin   s  cos   i sin  
 rs  cos  i sin   cos   i sin  

 rs cos cos   i cos sin   i sin  cos   i 2 sin  sin 

 rs   cos cos   sin  sin    i  cos sin   sin  cos   
 rs  cos(   )  i sin(   ) 
From this we can observe that
and
(vii)
z1z2  z1  z2
Arg  z1z2   Arg z1  Arg z2 .
r (cos θ  i sin θ)
z1

z2 s(cos α  i sin α)
(cos θ  i sin θ) (cos α  i sin α)
 r

s (cos α  i sin α) (cos α  i sin α)
2
 r ( cos  cos   i sin  sin2   i 2cos 2 sin   i sin  sin  )
s
cos   i sin 
cos  cos   sin  sin   i(sin  sin   cos  cos  )
 r
s
1
 r  cos(   )  i sin(   ) 
s
From this we can observe that
and
z
z1
 1
z2
z2
z
Arg  1   Arg z1  Arg z2 .
z
 2
PAGE 9 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Examples:
 


13.
Simplify 3 cos   i sin   4 cos   i sin  .
3
3
2
2
14.
z
z1  9 cos   i sin  and z2  3 cos   i sin  . Find 1 .
2
2
6
6
z2




PAGE 10 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
15.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Let z  r  cos  i sin   . Find expressions for z 2 , z 2  z  z 3, z 4, z 5 and z n.
N.B. an easier way is to realise that
we already know that:
z2  z  z  r2
and
Arg z 2  Arg z  Arg z      2
 z 2  r 2  cos 2  i sin 2 
Now try:
Page 98, Exercise 5 – questions 1(d, e, f, i) 2(b, d)
From example 12 we can deduce that z n  r n  cos n  i sin n  . This is not a comprehensive proof:
it can be proved by induction. This is called De Moivre’s Theorem.
PAGE 11 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
z n  r n  cos n  i sin n 
De Moivre’s Theorem
Examples:
16.
Let z  3  i .
Write z in polar form and find and expression for z 4 (remember that   arg z   ).
Argand diagram:
Polar form:
r=
Arg z =
z=
Therefore,
z4 
PAGE 12 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
17.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Treat z  1  i like an algebraic expression (e.g. 1  x ).
Use the binomial expansion to find and expression for z 5
(remember i 2  1, i3  i, i 4  1, i5  i ).
Use De Moivre’s Theorem to find an expression for z 5 in polar form and show the solution
on an Argand diagram.
Hence, by equating the real parts of each of the above expressions,
show that cos 3   1 .
4
2
 
PAGE 13 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
18.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
(from SQA AH 2005)
Let z  cos  i sin  .
(a)
Use the binomial expansion to express z 4 in the form u  iv , where u and v are
expressions involving sin  and cos  .
3
(b)
Use De Moivre’s theorem to write down a second expression for z 4 .
1
(c)
Using the results of (a) and (b), show that
cos 4  p cos 2   q sec2   r, where       ,
2
2
cos 2 
stating the values of p, q and r.
6
Now try:
Page 98, Exercise 5 – question 5
Page 101, Exercise 6 – questions 1a, 2(c, d), 3d, 4(h, i, j), 6, 7
PAGE 14 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Sets of Loci on the Complex Plane
Complex equations and inequations can be used to define a set of points in an Argand diagram.
Examples:
19.
| z |  4 (where z= x + i y) represents the shaded area shown below:
y
z 4
4
 x2  y 2  4

x2  y 2  16
Which is a circle with centre (0, 0) and radius 4.
x
4
z 4
z  4 is the circumference and enclosed are of the circle
with centre (0, 0) and radius 4
20.
Sketch the loci of the points given by z  1  2 .
21.
On an Argand diagram indicate the loci of the points represented by z  2  z  2i .
PAGE 15 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
nth Roots of Unity - Roots of a Complex Number




1
An extension of de Moivre’s Theorem, z1  r n  cos   2k  i sin   2k 
n
n


where k  0, 1, 2, 3...... n  1 .
z  r  cos  i sin  
De Moivre’s Theorem states that for
z n  r n  cos n  i sin n 
we get
or to be more precise
z n  r n  cos n  Arg z   i sin n  Arg z  
where Arg z    2k for k  Z .
Therefore,

1
1
z n  r n cos 1   2k   i sin 1  θ  2kπ 
n
n
Examples:
22.

for k  0, 1, 2, 3...... n  1 .





 

Simplify z = 4 81(cos  i sin ) =  81 cos  i sin

2
2
2
2 

=




1
4
1
81 4  cos 1   2k  i sin 1   2k  where k = 0, 1, 2, 3
4 2
4 2




= 3  cos
  84k   i sin   84k  
remember to adjust for      .


For k = 0,
z  3 cos   i sin 
8
8
For k =1,
z  3 cos 5  i sin 5
8
8
For k = 2,
z  3 cos 9  i sin 9  3 cos 7  i sin 7
8
8
8
8
For k = 3,
z  3 cos 13  i sin 13  3 cos 3  i sin 3
8
8
8
8



 

 
PAGE 16 OF 21


BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
23.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Solve for z when z 4  2  2i .
z 4  (2) 2  (2) 2  2 2
  

arg  z 4   tan 1 2       3 
2
4
4




 N.B. it's in the 3
rd


quadrant 

3
3
1

1

 cos  4  4  2k   i sin  4  4  2k  





3 

 2 8  cos  1  3  2k   i sin  1  3  2k   where k  0, 1, 2, 3
4
4
4

4


z 2 2
1
4

k = 0 gives,



   
   
3
z  2 8  cos  3  i sin  3 
16
16 

3
 2 8  cos 3  i sin 3 
16
16 

 
 
 
 
k = 1 gives,
3
z  2 8  cos 5  i sin 5 
16
16 

k = 2 gives,
3
z  2 8  cos 13  i sin 13 
16
16 

k = 3 gives,
3
z  2 8  cos

3
 2 8  cos

3
 2 8  cos

 2116   i sin  2116  
  1116   i sin   1116  
 1116   i sin  1116  
Now try:
Page 106, Exercise 7 – questions 1(b, c, d, g), 2
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BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Examples (from SQA AH exams 2001, 2002, 2005 and 2003):
24.
Verify that i is a solution of z 4  4 z 3  3z 2  4 z  2  0 .
Hence find all other solutions.
5
PAGE 18 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
25.
(a)
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Given that 1  cos   i sin  ,       , state the value of  .
1
(b)
Use de Moivre’s Theorem to find the non-real solutions, z1 and z2 ,
of the equation z 3  1  0 .
5
Hence show that z12   z2 and z22   z1 .
2
(c)
Plot all the solutions of z 3  1  0 on an Argand diagram and state their geoemtrical
significance.
3
PAGE 19 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
26.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Identify the locus in the complex plane given by z  1  2 .
3
27.
Given the equation z  2i z  8  7i , express z in the form a  ib .
4
PAGE 20 OF 21
BRIDGE OF DON ACADEMY – FACULTY OF MATHEMATICS
28.
ADVANCED HIGHER – UNIT 2: COMPLEX NUMBERS
Given that w  cos  i sin  , show that 1  cos   i sin  .
w
1
Use de Moivre’s theorem to prove wk  wk  2 cos k , where k is a natural
number.
3

Expand w  w1

4
by the binomial theorem and hence show that
cos4   1 cos 4  1 cos 2  3 .
8
2
8
5
PAGE 21 OF 21