Download Complex Numbers Syllabus

Complex Numbers Syllabus
Ordinary Level
Higher Level
Slide 1
THE ARGANG DIAGRAM
Slide 2
Slide 3

Complex numbers have both a real and an
imaginary part.

They are usually written as a+ib, where a is
the Real part and b is the imaginary part.

For the complex number 2-4i, the real part is
2 and the imaginary part is -4. (Remember
the sign belongs to the number that comes
after it)
Slide 4
Slide 5
Slide 6
Slide 7

Add 2+2i+(2-i)

Add the real parts 2+2=4

Add the imaginary parts 2+(-1)=2-1=1

2+2i+(2-i)= 4+i
Slide 8
Slide 9
Plotting B+2-i
Plotting D+2-i
1-i+2-i=
1+2-i-i=
3-2i
-4-2i+2-i=
-2-3i
Plotting C+2-i
-2+3i+2-i=
2i
Slide 10
Slide 11
Slide 12

All the points are shifted by the same
amount. This is a translation.

Moving the points by the same amount the
distance and direction from the origin to 2-i
Slide 13
Slide 14
When we subtract a
complex number we need
to make sure we subtract
both the real and imaginary
parts
2+2i-(2-i)
Real part 2-2=0
Im part 2i-(-i)=2i+i=3i
2+2i-(2-i)=3i
Slide 1
Slide 2
For complex numbers:

Remember:
( x  3)(2 x  4)
Breakup the first bracket
x(2 x  4)  3(2 x  4)
( 2  4i )(3  2i )
2(3  2i )  4i (3  2i )
6  4i  12i  8i 2
6  8i  8i 2
2 x 2  4 x  6 x  12
but i 
2 x 2  10 x  12
so i 2  (
1
 1 ) 2  1
6  8i  8( 1)
6  8i  8
14  8i
Slide 3

First we can look at multiplying by i
A = 3+2i
Multiplying by i
i(3+2i)
=3i+2i2
=3i +2(-1)
=-2+3i
A is rotated by
90 anti
clockwise
Slide 4
Multiplying A by –i
-i(3+2i)
=-3i-2i2
=-3i-2(-1)
=2-3i
Multiplying by i rotates
anti clockwise
Multiplying by –i rotates
clockwise
Slide 5




Multiplying by i rotates by
90 degrees
If I multiply again by i it
rotates another 90degrees
and so on.
multiplying by i² or -1
rotates by 180 degrees
Multiplying by i³ or –i
rotates the point by 270
degrees anti clockwise and
multiplying by i4 or by 1
rotates the point afull 360
degrees
Slide 6


Remember: the
modulus is the
distance from the
complex number to the
origin.
We can find the
modulus using the
formula
z  a 2  b2

Where a complex
number is a+ib
Multiply each point by 2 to H, I, J and K
Slide 7
(-2+i)(2-2i)
=-2(2-2i)+i(2-2i)
=-4+4i+2i-2i2
=-4+2+6i
=-2+6i
A is rotated by the same amount as
the rotation between the Real axis
and B and their moduli are multiplied
Slide 1
COMPLEX NUMBERS
MODULUS AND
COMPLEX CONJUGATE
Slide 2
Conjugate,
z  a  ib
Four points A, B, C,
and D are shown with
their Complex
conjugates A’, B’, C’
and D’.
The complex conjugate
is the image of the
point under axial
symmetry in the real
line.
To find the complex
conjugate of any
complex number I
change the sign of the
Imaginary part.
The complex conjugate
of 2-4i is 2+4i
Slide 3
The modulus
The modulus of A is distance from the origin to the point A.
Looking at the Modulus as the hypotenuse of a right angled triangle
we can use Pythagoras to the modulus.
h²=x²+y²
The x is the
distance along
the real axis.
The y is the
distance along
the imaginary
axis
h²=4²+3²
h²=16+9
h²=25
h= 5
Slide 4
Modulus Formula

For a complex number z=a+ib, it’s modulus
Z  a 2  b2
Eg. z = 2-4i
z 
2 2  ( 4 ) 2

4  16

20

4 5
2 2
Slide 1
Slide 2

We start with a number like
5  5i
2i


We need to get rid of the 2-i beneath the line
We can do this by multiplying by the complex
conjugate of 2-i; 2+i on both the top and the
bottom
Slide 3
(5  5i ) (2  i )
(2  i ) (2  i )
5(2  i )  5i (2  i )

2( 2  i )  i ( 2  i )
10  5i  10i  5i 2
4  2i  2i  i 2
10  15i  5(1)

4  (1)
10  15i  5

4 1
5  15i

5
5 15
  i
5 5
 1  3i

Slide 4
Slide 1
Complex Root
Slide 2

Complex roots of a quadratic equation

3z²+4z+3=0
Use the formula

 b  b 2  4ac
2a

a=3, b=4, c=3
Complex roots
Slide 3

Fill in to the formula
4
4 2  4(3)(3)
2(3)
 4  16  36
6
 4   20
6
 4  20  1

6
 4  4 5 1

6
 4  2 5i
 4  2 5i

or 
6
6
2
5
2
5
 
i or   
i
3
3
3
3


Complex roots
Higher Level Course
Polar Form
When using the polar form we will need to change our calculator into radians. CASIO(Shift, Setup, 4)
and change back once you’re finished.
Instead of placing a point in space based on its x and y coordinates (Cartesian coordinates). We can
also place a point based on its angle from the x-axis and its distance from the origin and write the
point in the form
, where r is the distance form the origin and the angle made
with the positive x-axis.
The distance from the origin can be found by getting the modulus of the complex number.
The angle made with the positive x-axis is called the argument(arg). We can use trigonometry to
find the angle but we need to look at the quadrant our point is in to decide if my angle is positive
or negative and if the angle is acute or obtuse. Any point in the first two quadrants or with positive y
coordinates will give us a positive angle. While any below the x-axis or in the last two quadrants will
be negative. The angles in the 2nd and 3rd quadrants (negative x coordinate) will have an obtuse and
angle and the angle we find using trigonometry will need to be subtracted from π.
First Quadrant
Point
3+i
We can use trigonometry to find the arg
The Argument is
r is the modulus r=6
Second Quadrant
Point
-3+i
We can use trigonometry to find the arg
The Argument is
r is the modulus r=6
because
Third Quadrant
Point
- 3-i
The Argument is
But the arg is also negative because it is
under the x-axis so
r is the modulus r=6
)
This can also be written as
See page 13 in formula and tables book
Fourth Quadrant
Point
3-i
The Argument is
But the arg is also negative because it is
under the x-axis so
r is the modulus r=6
)
This can also be written as
We use the polar form because it allows us to multiply, divide and get powers of complex numbers
quickly. This is because multiplication of complex involves adding the angles and multiplying the
moduli (as seen above).
If z1= r1[Cos 1++iSin 1] and z2= r2[Cos 2+iSin 2]
z1z2=r1r2[Cos
1+ 2)+iSin
1+ 2)]
and
Examples:
[2(Cos ++iSin )][3(Cos +iSin )]= (2)(3)[Cos
+
)+iSin
=6[Cos
)+iSin
)]
=3[Cos
)+iSin
)]
+
)]
De Moivre’s Theorem ,nЄℕ
If z= r [Cos +iSin ] then zn= rn [Cos
+iSin
]
Example
Use de Moivre’s theorem to write the following in the form a+ib
To put into a+ib form put
And
to get the Im(z)
Ans =15625+i0
=15625
nto calculator(in rad mode) to get Re(z)
You can also be asked to prove identities using deMoivre and expansions
Example
Prove
Using deMoivre
For z=
z4=Cos
+iSin
but z4=
Putting the Re(z) parts equal to each other
Since
and
We have been able to replace all any SinA with CosA
De Moivre can also be used to find the square root, cube root, etc.
Using the fact
that
or
. However, if we use the usual polar form we get only 1 possible root and
not the n roots we should be able to find. To get over this problem we use the general polar form
which allows us to add any number of full rotations by adding any number of 2πn.
The general polar form
Solve the following
z4=-4-4i
Change into the general polar form of ω=-4-4i
Arg
But it is in the third quadrant so the arg will be negative and I need to subtract it form π
Arg=
The general polar form of ω=
z4=ω so z=
When n=0
When n=1
When n=2
When n=3
When n=4
This is what the fifth roots of -4-4i look like when plotted. Note how all the points have the same
modulus and are all the same angle from each other. The fifth roots are all
each other. You can be asked to sketch this.
Proof of De Moivre’s Theorem by Induction for nЄℕ.
Induction involves three steps.
1. Show true for n=1
2. Assume true for n=k
3. Prove true for n=k+1
Show that If z=r [Cos +iSin ] then zn= rn [Cos
If n=1 then z1= r1 [Cos
+iSin
]
z=r [Cos +iSin ]
Assume true for n=k, the
If z= r [Cos +iSin ] then zk= rk [Cos
+iSin
Show true for n=k+1
zk+1= r [Cos +iSin ] rk [Cos
+iSin
]
]
+iSin
]
away from
= r rk [Cos +iSin ] [Cos
+iSin
]
=rk+1 [Cos +iSin ] [Cos
+iSin
]
=rk+1 [Cos
+iCos
=rk+1 [Cos
+Cos
Sin +i2 Sin Sin
Sin -Sin Sin
]
]
=rk+1 [Cos
+Cos
Sin ]
=rk+1 [Cos
+Cos
Sin ]
From page 14 table and formula book
Cos(A+B)= cosAcosB-sinAsinB
Sin(A+B)= sinAcosB+cosAsinB
=rk+1 [Cos
+
=rk+1 [Cos
+ ]
]
QED
Finding the complex roots of functions
In order to find the complex roots of function it is easier to use the conjugate root theorem.
This states that if
is a root then
is also a root. First lets look at
what happens when we add and multiple complex numbers and their conjugates.
P(z) is a complex polynomial of the form
P(z)=
+
+
+......+
Where z=a+ib and a, b Є R
P(z)=0 when z is a root by the factor theorem.
P(z)=0 then
=0
P( )=
+
+
+......+
P( )=
because
+
+
+......+
P( )=
+
+
+......+
since each a is a constant
P( )=
Because
P( )=
Therefore, P( )=0 and from the factor theorem
.
This means for all z a factor of a polynomial then it’s conjugate is also a factor.
Example
-2,z1,z2 are the roots of the cubic equation z3+8=0
z2  2z  4
z  2 z3  0z 2  0z  8
-( z 3  2 z 2 )
-2z 2  0 z
-(-2z 2  4 z )
4z+8
(4 z  8)
0
Use
b  b 2  4ac
2a
on to get the other two factors z 2  2 z  4
(2)  (2) 2  4(4)
2
2  4  16
2
2  12
2
2  i2 3
2
1  i 3 or 1  i 3