Download complex numbers in polar

The Complex
Plane;
DeMoivre's
Theoremconverting to
rigonometric form
z  x  yi
We can take complex numbers given as
and convert them to polar form. Recall the conversions:
x  r cos
y  r sin  z  r cos   r sin  i
Imaginary
Axis
factor r out
 r cos  i sin  
The magnitude or modulus of
z is the same as r.
Real
Axis
Do Now:
Plot the complex
number:
z   3 i
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
2
 4 2
The angle 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

 principal  
6
6
5
5 

z  2 cos
 i sin

6
6 

5
or "2cis
"
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
Convert the following:
1) 3  3i
2)  4i
11
11
3) 4(cos
 i sin
)
6
6
Do Now: Convert the following
to trigonometric form:
3 1
z1 
 i
2 2
z2  1 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 r values and Add the Angles
 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 r values and add the angles)
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 r values and subtract the angles)
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 
z1z2  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 sin120 ,
o
find: (a) zw
o
(b) z w
o
o




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 r values





add the angles
(the i sine term will have same angle)
 24  cos160  i sin160
If you want the answer
in complex
coordinates simply
compute the trig
functions and multiply
the 24 through.




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 r values





add the angles
(the i sine term will have same angle)
 24  cos160  i sin160
 24 0.93969  0.34202i 
 22.55  8.21i
If you want the answer
in complex
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 r values
 
2
 cos  80
3

subtract the angles
 i sin 80 
In polar form we want an angle between 0° and 180°
“PRINCIPAL ARGUMENT”
2
o
o 
 cos 100  isin 100 
3




4 cos 40  i sin 40
z

w 6 cos120  i sin 120
 


4
 cos 40  120  i sin 40  120
6
divide the r values
 
2
 cos  80
3

subtract the angles
 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
Today’s theorem is used to raise
complex numbers to powers.
Do Now: simplify (foil)
 
 3 i
2
Today’s theorem is used to raise complex
numbers to powers.
Do Now: simplify
 3 i  2 2i 3
2
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
r value to that power and multiply the angle by that power.
This theorem is used to raise complex numbers
to powers.
2
 3  i
Instead let's convert to polar form
and use DeMoivre's Theorem:
CONVERT:
r   3  1  4  2
2
2
 1 
  tan 

 3
1
APPLY DEMOIVRE’S THEOREM:
z  2cis150
z  2 cis(2 150)  4cis300
2
2
CONVERT BACK:
4(cos 300  i sin 300)  2  2i 3
 1
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.
 3 i
4
you would need to FOIL
and multiply all of these
together and simplify
powers of i --- UGH!
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
 3
  tan 1 
5
  150 
6
o
5
z  2cis
6




5

5

4
Z  2 cos isin 
4
This theorem is used to raise complex numbers
to powers. It would be a lot of work to find
  5
5 
Z  2  cos  isin 
6
6 
 
4
4
 3 i
  5 
 5 
 2 cos 4    i sin  4  
 6 
  6
  20   20 
16 cos   isin  
  6   6 
4
 1 
 
3
i 
 16     
 
 2
2

 

 8  8 3i
4
Example:
Remember:
1
3
 i

2 
2
3
Convert
Apply Demoivre
Convert back
1st : (Put into calculator and see what the value is)
Example:
Convert:
1
3
3
 i
  (1cis(60))
2 
2
3
Example:
Convert:
Apply:
1
3
3
 i
  (1cis(60))
2 
2
3
1 cis(3 60)  1cis(180)
3
Convert back:
1

3
 i
 
2 
2
1cis(180)  1cis(180)
3
1(cos(180)  i sin(180))  1 0i  1
Solve the following over the set of complex numbers:
z 1
3
We get 1 but we know that there are 3
roots. So we want the complex cube
roots of 1.
Using DeMoivre's Theorem , we can
develop a method for finding complex
roots. This leads to the following formula:







360k

360k
n
zk  r cos   isin   
  n n   n n 
where k  0, 1, 2, , n  1
N is the root and k goes from 0 to 1 less
than the root.
We find the number of roots necessary







360k

360k
n
zk  r cos   isin  
  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)
2
2
1  0 
  tan    0 
r  1  0  1
1
1cis0
  0 360k 
 0 360k 
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







360k

360k
n
zk  r cos   isin   
  n n   n n 
  0 360k 
 0 360k 
zk  1 cos  
  i sin  
, for k  0, 1, 2
3
3 
3 
 3
zk  1cos 0  i sin 0
3
  0 360 0 



360
0
0


z0  3 1 cos  
  i sin  

3 
3 
3
 3
  0 360 1



360
1
0


3
z1  1 cos  
  isin  

3 
3 
3
 3
  0 360 2



360
2
0


z2  3 1 cos  
  isin  

3 
3 
3
 3
  0 360k 
 0 360k 
zk  1 cos  
  i sin  
, for k  0, 1, 2
3
3 
3 
 3
zk  1cos 0  i sin 0
3
  0 360 0 
 0 360 0 
z0  1 cos  
  i sin  

3
3
3
3



 
3
 1cos0  i sin 0  1
  0 360 1  0 360 1
z1  1cos  
 isin  

  3 3   3 3 
1
3


 1cos 120   i sin 120    
i
2
2
  0 360 2  0 360 2
z2  3 1 cos  
 isin  

  3 3   3 3 
1
3


 1cos 240  i sin 240   
i
2 2
3
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.
Find the cube roots of:

1 i 3

Find z in trig form:
Find the cube roots of:
1  i 3
1
3
z  2cis(60)
 60  3
zk  2 cis 
  2 cis(20)
 3 
1
3
the cube roots in trig form:
1 i 3
1
3
Root 1
Root 2
Root 3
z0  3 2 cis(20)

360  3
z1  2 cis 20 
  2 cis100

3 

720  3
3
z2  2 cis 20 
  2 cis220

3 
3
the cube roots in complex form:
1 i 3
1
3
z0  2 cis(20)  1.18  .43i
3

360  3
z1  2 cis 20 
  2 cis100  .22 1.24i

3 

720  3
3
z2  2 cis 20 
  2 cis220  .97  .81i

3 
3
Find all real and complex cube roots of -8
Z= -8 + 0i
Find trig form first:
Find all real and complex roots of -8
Z= -8 + 0i
Find trig form first:
r8
  180
z  8cis180
 180 
3
zk  8 cis 
  2cis60
 3 
Now find the 3 roots in trig form
Find all real and complex roots of -8
Z= -8 + 0i
 180 
3
zk  8 cis 
  2cis60
 3 
the 3 roots in trigonometric form:
z0  2cis60

360 
z1  2cis  60 
  2cis180

3 

720 
z2  2cis  60 
  2cis300
Find all real and complex roots of -8
Z= -8 + 0i
 180 
3
zk  8 cis 
  2cis60
 3 
the 3 roots in complex form:
z0  2cis60  1 i 3

360 
z1  2cis  60 
  2cis180  2

3 

720 
z3  2cis  60 
  2cis300  1  i 3
Do Now:
Find all real roots of
4
16
Find all real and complex fourth roots of 16
Z= 16+ 0i
Find trig form first:
16cis0
1
4
1
4
0
zk  (16cis0)  16 cis  2cis0
4
Write it 4 times:
Find all real and complex fourth roots of 16
Z= 16+ 0i
1
4
1
4
0
zk  (16cis0)  16 cis  2cis0
4
z0  2cis0
360
z1  2cis(0 
)  2cis90
4
720
z2  2cis(0 
)  2cis180
4
1080
z3  2cis(0 
)  2cis270
Find all real and complex fourth roots of 16
Z= 16+ 0i
1
4
1
4
0
zk  (16cis0)  16 cis  2cis0
4
2cis0  2
360
2cis(0 
)  2cis90  2i
4
720
2cis(0 
)  2cis180  2
4
1080
2cis(0 
)  2cis270  2i
4
Find all real and complex 6th roots of
Z= 1- i
Find all real and complex 6th roots of
Z= 1- i
z  2cis(45)
45 12
zk  (2 ) cis
 2cis(7.5)
6
1 1
2 6
Find all real and complex 6th roots of 1- i
z0  12 2 cis(7.5)

360  12
12
z1  2 cis 7.5 
  2 cis52.5  .64  .84i

6 

720  12
12
z2  2 cis 7.5 
  2 cis112.5

6 

1080  12
z3  2 cis 7.5 
  2 cis172.5

6 

1440  12
12
z4  2 cis 7.5 
  2 cis232.5

6 

1800  12
12
z5  2 cis 7.5 
  2 cis292.5

6 
12