Transcript Lesson 6.6 De Moivre`s Theorem and nth roots

Start Up Day 54
1. PLOT the complex number, z = -4 +4i
2. Determine the exact value of the absolute
value of z. (The distance from (0, 0))
3. Determine the angle Θ, in standard
position, of the rotation to this complex
number “z”.
4. Determine the cosine and sine of Θ.
OBJECTIVE: SWBAT Plot and find absolute values of
complex numbers, Write the trigonometric forms of
complex numbers & Multiply and divide complex numbers
written in trigonometric form. SWBAT Use DeMoivre’s
Theorem to find powers & nth roots of complex numbers.
EQ: How are complex numbers represented graphically?
How is the absolute value of a complex number determined?
How can DeMoivre’s theorem be used to express powers
and/or roots of complex numbers?
HOME LEARNING:p.511‐513 # 2, 4, 15, 16, 29, 30, 32, 33,
35
Lesson 6-6
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
z  x  yi
z

x
2
y
The modulus or absolute
value 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
Real
of z, with requirement that 0 
Axis
 < 2.
modified for the
correct quadrant
1  y 
  tan   (always between 0
x
and 2)
z  x  yi
We can take complex numbers given as
and convert them to trigonometric (or polar) form.
y  r sin 
x  r cos
Imaginary
Axis
z =r
1

 3

x
factor r out
y
z  r cos   r sin  i
 r cos  i sin  
The modulus of z is the same
as r.
Plot the complex number: z   3  i
Real
Axis Find the trigonometric or
polar form of this number.
r
 3   1
2
2
 4 2
5
5    tan 1  1  With an

adjustment for
z  2 cos
 i sin


3


6
6
Quad II   5


6
Use the inverse tangent—carefully-- to determine the
angle! When in Doubt, Sketch it out!
Imaginary
Axis
 1 
  tan 
 but in Quad II
 3
1
z =r
1

 3

x
y
Real
Axis
5
5 

z  2 cos
 i sin

6
6 

5

6
5p
arg z =
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


i
sin
cos


i
sin

1 as to what
2
2
see if you can make a1 statement
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 z 2  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 z 2  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 z 2  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 z 2  0, then
z1 r1
 cos1  2   i sin1  2 
z2 r2
z1 r1
 cis 1   2 
z 2 r2




If z  4 cos 40  i sin 40 and w  6 cos 120  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:
Abraham de Moivre
(1667 - 1754)
DeMoivre’s Theorem
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   2 4 cos 4  5   i sin  4  5 
6
6 
6 
6 

 
 
4
  10 
 10 
 16cos
  i sin 

 3 
  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:
   2k 
  2k 
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 2 k 
 0 2 k
z k  1 cos 
  i sin  
3 
3
3
 3
3

, for k  0, 1, 2

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.
   2k
z k  r cos 
n
 n
n
We want cube
root so use 3
numbers here

  2k
  i sin  
n

n



  0 2k 
 0 2k
zk  1 cos 
  i sin  
3 
3
3
 3
3
  0 20
z0  1cos 
3
 3
3

 0 20
  i sin  
3

3

, for k  0, 1, 2


 1cos 0   i sin 0   1
 Here's the root we
  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 
z 2  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:1,
Let's plot these on the complex
plane
each line is 1/2 unit
1
3
1
3
 
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.
VIDEOS:
http://youtu.be/MO6
qU3SCLbM
http://www.youtube.
com/watch?feature=
player_embedded&v
=MO6qU3SCLbM
http://youtu.be/tAIxdEVu
TZ8
On a side note, you will
sometimes see the
trigonometric form of a
complex number rewritten is
a more condensed format:
For example-2(cos 30° + i sin 30°) can
also be noted as:
2cis30°
*Another point of interest is
that r (cosθ+ isinθ) = reiθ