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CIE Centre A-level
Pure Maths
P3 Chapter 16
© Adam Gibson
COMPLEX NUMBERS
Girolamo Cardano:
1501-1576
A colourful life!
ax  bx  cx  d
3
2
x  px  q  0
3
The “depressed cubic”;
other cubics can be expressed
in this form.
COMPLEX NUMBERS
Cardano “stole” the method from
Tartaglia. Finding x you must
write:
2
3
q
q
p
p
a
3
2
u


, x
u 
2
4 27
3u
3
can be less than zero
but the solution
is a real number!
Square roots of negative numbers can be useful,
just as negative numbers themselves.
COMPLEX NUMBERS
By introducing one number
i  2 1
we can solve lots of new problems, and make
other problems easier.
It can be multiplied, divided, added etc. just as any
other number; but the equation above is the only
extra rule that allows you to convert between i and
real numbers.
Start by noticing that
4  4  1  2i
So the square root of
any negative number
can be expressed in terms
of i.
COMPLEX NUMBERS
Therefore we can solve equations like:
x 2  8
But what is
The answer is
x  2 2i
1  3i  1  3i
The two types of number cannot be “mixed”.
Numbers of the form
k i , k 
are called imaginary numbers (or “pure imaginary”)
Numbers like 1, 2, -3.8 that we used before are called
real numbers.
When we combine them together in a sum we have
complex numbers.
COMPLEX NUMBERS
COMPLEX NUMBERS
To summarize,
z  a  bi
•a and b are real numbers
•a is the “real part” of z; Re(z)
•b is the “imaginary part” of z; Im(z)
•The sum of the two parts is called a
“complex number”
COMPLEX NUMBERS
Adding and subtracting complex numbers:
z1  (2  3i)
z2  (4  9i )
z1  z2  6  6i
(a  bi)  (c  di)  (a  c)  (b  d )i
For addition and subtraction the real and imaginary
parts are kept separate.
COMPLEX NUMBERS
Multiplying and dividing complex numbers:
z1  (2  3i)
z2  (4  9i )
z1 z2  (2  3i)  (4  9i)
 2  4  (2  9i)  (3i  4)  (3i  9i)
 8  18i  12i  (27  i 2 )
 35  6i
(a  bi )  (c  di )  (ac  bd )  (bc  ad )i
Notice how, for multiplication, the real and imaginary
parts “mix” through the formula i2 = -1.
COMPLEX NUMBERS
Multiplying and dividing complex numbers:
z1  (2  3i)
z2  (4  9i )
z1
(2  3i)

z2
(4  9i)
Remember
this trick!!
(2  3i) (4  9i)


(4  9i) (4  9i)
8  18i  12i  (27  i 2 )

4  4  36i  36i  (9  9  i 2 )
19  30i
19 30
Read through Sections 
  i
97
97 97
16.1 and 16.2 to make
sure you understand the basics.
COMPLEX CONJUGATES
Now that we have introduced complex numbers, we can
view the quadratic solution differently.
b  b2  4ac
x
2a
Now there are always two solutions, albeit they can be
repeated real solutions.
If the equation has no real roots, it must have two
complex roots.
what is the other?
If one complex root is 1  8i
1  8i
These two numbers are called
“complex conjugates”.
COMPLEX CONJUGATES
What are the solutions to
?
x 2  6 x  21  0
3  2 3i
* means conjugate
If we write
z  3  2 3i
Then the complex conjugate is written as
z*  3  2 3i
Calculate the following:
z  z*
 6  2 Re( z )
z  z*
 4 3i  2 Im( z )
zz
*

3  2 3
2

2
 21
This will be discussed
later.
 z
2
COMPLEX POWERS
What happens if we
square a complex number z?
And then square
its conjugate, z*:
z  x  iy
z *  x  iy
z 2  ( x  iy )( x  iy )
 x 2  xiy  xiy  y 2
z 
 ( x  y )  i (2 xy )
 x 2  y 2  i (2 xy )
2
* 2
2
 ( x  iy )( x  iy )
Compare the two results;
they are complex conjugates!
z   z 
2 *
* 2
Later we will understand
this result geometrically
COMPLEX POWERS
Continuing these investigations further (you may study
in your own time if you wish):
  
*
*
*
z

z

z

z
 1 2 1 2
z
n *
 z
*
n
This will be easy to justify
later.
Examine the argument on page 228.
This is a key idea, although you don’t
have to understand the proof.
Non-real roots of polynomials
with real coefficients
always occur in conjugate pairs.
COMPLEX POWERS
Tasks
Find all the roots of the following two polynomials:
a) z 3  z 2  7 z  65
b) z  z  1
4
2
The first example can be attacked using the factor
theorem.
Examining +/-1,+/-5,+/-13 gives one root as -5.
Equating coefficients therefore gives:
z 3  z 2  7 z  65  ( z  5)( z 2  4 z  13)
COMPLEX POWERS
 z 3  z 2  7 z  65  ( z  5)( z  2  3i)( z  2  3i)
The roots are therefore -5, 2-3i, 2+3i.
b) z 4  z 2  1
The second example looks simpler but is, in a way,
more difficult.
First set w = z2.
w2  w  1  0
It seems we have
to find the square
root of a complex
number!
1  3
w
2
1 3i
1 3i
2
2
z  
or z   
2 2
2 2
COMPLEX POWERS
Algebra is not the best way to do it, but let’s try anyway.
z  a  ib
 (a  ib)(a  ib)  z
Re :
a b  x
Im :
2ab  y
2
2
The next step is important to
understand. It is called
“equating real and imaginary
parts”.
y
a
2b
y
2
b  x
2
4b
Simultaneous
equations.
Let’s apply it
to our problem.
COMPLEX POWERS
1 3i
z  
2 2
2
1 3i
( x  iy )( x  iy )   
2 2
1
3
2
2
x  y   , 2 xy 
2
2
9
1
2
x 

2
16 x
2
16 x 4  8 x 2  9  0
8  24
1 3
1
x 
    1 or
32
4 4
2
2
COMPLEX CONJUGATES
Special properties of complex conjugates:
Im( z )
z
z  z*  2 Re( z )
*
z  z  2 Im( z )

Re( z )

z*
What is
zz *
?
How do we know
it must be a real
number?
COMPLEX CONJUGATES
z  r (cos   i sin  )
z*  r (cos   i sin   )
arg( z1 z2 )  arg( z1 )  arg( z2 )
 arg( zz* )      0
zz*  z z*  z
zz *  z
2
2
or r 2
This is a very important result.
COMPLEX CONJUGATES
How many complex roots do the following polynomials
have?
A z10  3  4 z
10
B 65  63 z  z 2  z 3
3
C 3 z  4 z  18 z  13
5
2
5
4
See page 229. We always have n roots for a
polynomial of degree n. If the coefficients are
real numbers, then we also know that any nonreal roots occur in complex conjugate pairs.
If 1-8i is a root of polynomial B, what are the other roots?
POLAR COORDINATE FORM
Im( z )
z
r
r sin 

Re( z )
r cos
The modulus is the length of the line
from 0+0i to the number z, i.e. r.
The argument is the angle between the
positive real axis and that line, by
convention we use     
POLAR COORDINATE FORM
Find, to 3 s.f. the modulus and argument of the following
complex numbers:
4i
7  6i
arg( z )  
r  42  12  17
modulus
z  r (cos   i sin  )
To find θ we have two equations:
4
1
cos  
, sin  
17
17
  0.245
POLAR COORDINATE FORM
4i
7  6i
z  r  62  72  17 5  9.22 (3 s.f.)
7
6
cos  
, sin  
17 5
17 5
arg( z )    2.43 (3 s.f.)
 z  7  6i  9.22(cos(2.43)  i sin(2.43))
EXPONENTIAL FORM
y ( x)  cos( x)  i sin( x)
or
z ( )  cos( )  i sin( )
dy
  sin x  i cos x
dx
 i (i sin x  cos x)
 iy
Which function does this?
dy
 ky
dx
EXPONENTIAL FORM
y  Ae
kx
So (not proof but good enough!)
z  r (cos   i sin  )  re
i
If you find this incredible or bizarre, it means you
are paying attention.
Substituting
i
 
e 1  0
gives “Euler’s jewel”:
which connects, simply, the 5 most
important numbers in mathematics.
EXPONENTIAL FORM
We can write any complex number in this form re i
As before, r is the modulus and θ is the argument.
Examples:
i 2
i
2i 
e
6
6ei 0
2e
1  3i  2e
 i 2
Do you see how easy it
is to calculate powers?

Find 1  3i

10
 i 3
1  i 23
e
1024