Transcript Document

The Complex
Plane;
DeMoivre's
Theorem
Remember a complex number has a real part and an
imaginary part. These are used to plot complex
numbers on a complex plane.
z  x  yi
z  x y
2
Imaginary
Axis
2
z  x  yi
z

x
y
Real
Axis
The magnitude or modulus
of z denoted by z is the
distance from the origin to
the point (x, y).
The angle formed from the
real axis and a line from the
origin to (x, y) is called the
argument of z, with
requirement that 0   < 2.
 y
  tan  
x
1
modified for quadrant
and so that it is
between 0 and 2
z  x  yi
We can take complex numbers given as
and convert them to polar form. Recall the conversions:
Imaginary
Axis
z =r
1

 3
z  r cos   r sin  i
y  r sin 
x  r cos 

x
factor r out
 r cos  i sin  
The magnitude or modulus of
z is the same as r.
y
Plot the complex number: z   3  i
Real
Axis Find the polar form of this
number.
r
 3   1
2
5
5    tan1  1 

z  2 cos  i sin

 3

6
6 

2
 4 2
but in Quad II
5

6
The Principal Argument is between - and 
Imaginary
Axis
 1 
  tan 
 but in Quad II
 3
1
z =r
1

 3

x
y
Real
Axis
5

6
5
5
arg z 
 principal arg 
6
6
5
5 

z  2 cos  i sin

6
6 

It is easy to convert from polar to rectangular form
because you just work the trig functions and distribute
the r through.
5
5   3 1 

z  2 cos  i sin
 i    3  i
  2 
6
6   2 2 

1
2
3

2
1
2
 3
5
6
If asked to plot the point and it
is in polar form, you would
plot the angle and radius.
Notice that is the same as
plotting
 3 i
Let's try multiplying two complex numbers in polar
form together.
z1  r1 cos1  i sin 1 
z2  r2 cos 2  i sin  2 
z1 z2   r1  cos 1  i sin 1    r2  cos  2  i sin  2  



Look at where
andwhere
we
ended
up and
 r1r2 we
cosstarted
1  i sin
cos


i
sin

1
2
2
see if you can make a statement
as to what
happens
to
Must
FOIL
these two complex
the r 's and the  's when
you
multiply
 numbers.
r1r2 cos 1 cos  2  i sin  2 cos 1  i sin 1 cos  2  i 2 sin 1 sin  2

Replace i 2 with -1 and group real terms and then imaginary terms
Multiply the Moduli and Add the Arguments
 r1r2 cos1 cos2  sin 1 sin 2   sin 1 cos2  cos1 sin 2 i
use sum formula for cos
use sum formula for sin
 r1r2 cos1  2   i sin1  2 

Let z1  r1 cos 1  i sin 1  and z2  r2 cos  2  i sin  2 
be two complex numbers. Then
z1 z2  r1r2 cos1  2   i sin1  2 
(This says to multiply two complex numbers in polar
form, multiply the moduli and add the arguments)
If z2  0, then
z1 r1
 cos1  2   i sin1  2 
z2 r2
(This says to divide two complex numbers in polar form,
divide the moduli and subtract the arguments)
Let z1  r1 cos 1  i sin 1  and z2  r2 cos  2  i sin  2 
be two complex numbers. Then
z1 z2  r1r2 cos1  2   i sin1  2 
z1 z2  r1r2 cis1  2 
If z2  0, then
z1 r1
 cos1  2   i sin1  2 
z2 r2
z1 r1
 cis 1   2 
z2 r2




If z  4 cos 40  i sin 40 and w  6 cos120  i sin 120 ,
find : (a) zw
(b) z w





 

zw  4 cos 40  i sin 40  6 cos120  i sin120 

 


 4  6 cos 40  120  i sin 40  120
multiply the moduli





add the arguments
(the i sine term will have same argument)
 24  cos160  i sin160
 24 0.93969  0.34202i 
 22.55  8.21i
If you want the answer
in rectangular
coordinates simply
compute the trig
functions and multiply
the 24 through.




4 cos 40  i sin 40
z

w 6 cos120  i sin 120
 


4
 cos 40  120  i sin 40  120
6
divide the moduli
 
2
 cos  80
3

subtract the arguments
 i sin 80 
In polar form we want an angle between 0° and 180°
PRINCIPAL ARGUMENT
In rectangular
coordinates:
2
 0.1736  0.9848i   0.12  0.66i
3
You can repeat this process raising
complex numbers to powers. Abraham
DeMoivre did this and proved the
following theorem:
DeMoivre’s Theorem
Abraham de Moivre
(1667 - 1754)
If z  rcos  i sin is a complex number,
then
z  r cos n  i sin n 
n
n
where n  1 is a positive integer.
This says to raise a complex number to a power, raise the
modulus to that power and multiply the argument by that
power.
This theorem is used to raise complex numbers
to powers. It would be a lot of work to find




  3 i  3 i  3 i  3 i

Instead let's convert to polar form
and use DeMoivre's Theorem.
r

 3
2
1  4  2
2
 3  i
4
you would need to FOIL
and multiply all of these
together and simplify
powers of i --- UGH!
 1 
 but in Quad II   5
 3
6
  tan1 

4
 
5
5  
 3  i   2  cos
 i sin
   24 cos 4  5   i sin  4  5 
6
6 
6 
6 

 
 
4
  10 
 16  cos 
  3

 10 
  i sin 

 3



 1 
 
3
i 
 16     
 
 2
2

 

  8  8 3i
Solve the following over the set of complex numbers:
z 1
3
We know that if we cube root both sides we
could get 1 but we know that there are 3
roots. So we want the complex cube roots of
1.
Using DeMoivre's Theorem with the power being a
rational exponent (and therefore meaning a root), we can
develop a method for finding complex roots. This leads
to the following formula:
   2 k 
  2 k  
z k  r  cos  
  i sin  

n 
n 
n
 n
n
where k  0, 1, 2, , n  1
Let's try this on our problem. We want the cube roots of 1.
We want cube root so our n = 3. Can you convert 1 to
polar form? (hint: 1 = 1 + 0i)
1  0 
2
2
  tan    0
r  1  0  1
1
  0 2k 
 0 2k 
zk  1cos 
  i sin  
, for k  0, 1, 2
3 
3 
3
 3
3
Once we build the formula, we use it first
with k = 0 and get one root, then with k = 1
to get the second root and finally with k = 2
for last root.
We want cube
root so use 3
numbers here
   2 k 
  2 k  
z k  r  cos  
  i sin  

n 
n 
n
 n
n
  0 2k 
 0 2k 
zk  1cos 
  i sin 
, for k  0, 1, 2
3 
3 
3
 3
3
  0 20 
 0 20 
z0  1cos 
  i sin 
 1cos0  i sin 0  1
3 
3  Here's the root we
3
 3
3
  0 21 
 0 21 
z1  1cos 
  i sin  

3 
3 
3
 3
  2 
1
3
 2 
 1cos

i
sin



i



2 2
 3 
  3 
 0 22 
 0 22 
3 
z2  1cos 
  i sin  

3 
3 
3
 3
already knew.
3
  4 
 4
 1cos
  i sin 
 3
  3 
1
3

i
   
2 2

If you cube any of
these numbers
you get 1.
(Try it and see!)
We found the cube roots of 1 were:
Let's plot these on the complex
plane
each line is 1/2 unit
1
3
1
3
1,  
i,  
i
2 2
2 2
about 0.9
Notice each of
the complex
roots has the
same magnitude
(1). Also the
three points are
evenly spaced
on a circle. This
will always be
true of complex
roots.
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au