Transcript Trigonometric Form of a Complex Number

Trigonometric Form of a Complex
Number
Complex Numbers
Recall that a complex
number has a real component
and an imaginary component.
bi
z = 3 – 2i
The absolute value of a complex
number is its distance from the
origin.
z  a b
Imaginary
axis
a
z = a + bi
2
Argand Diagram
Real
axis
z = 3 – 2i
The names and letters are changing, but this
sure looks familiar.
2
z  32   2   9  4  13
2
The Trig form of a Complex Number
x a

r r
y b
sin   
r r
cos  
The trig form of the complex number z  a  bi
is z  (r cos  ri sin  )  r  cos  i sin   .
r is called the modulus and is the distance from
the origin to the point. r  a 2  b2
 is called the argument and is the angle
formed with the x-axis.
b
  tan 1  
a
r  a2  b2
b  r sin 

a  r cos 
How is it Different?
In a rectangular system, you go
left or right and up or down.
z  2  2i
In a trigonometric or polar system, you
have a direction to travel and a distance
to travel in that direction.
z  2  cos 45  i sin 45
Polar form (2,45)
Converting from Rectangular form to
Trig form
1. Find r. r  a 2  b2
b
2. Find  .   tan  
a
3. Fill in the blanks in z  r  cos   i sin  
1
1. Find r
r  42  32  16  9
r  25  5
2. Find 
  tan 1
3
 36.9
4
Convert z = 4 + 3i to trig form.
3. Fill in the blanks
z  5  cos 36.9  i sin 36.9 
Polar form  5,36.9 
Converting from Trig Form to
Rectangular Form
This one’s easy.
1.
Evaluate the sin and cos.
2.
Distribute in r
1. Evaluate the sin and cos
2. Distribute the 4.
Convert 4(cos 30 + i sin 30) to
rectangular form.
 3
1
4
i 
2
 2
2 3  2i
Multiplying Complex Numbers
To multiply complex numbers in
rectangular form, you would FOIL
and convert i2 into –1.
To multiply complex numbers in trig
form, you simply multiply the rs and
add the thetas.
 a  bi  c  di 
 r1  cos 1  i sin 1    r2  cos 2  i sin 2 
r1r2  cos 1  2   i sin 1  2  
ac  adi  bci  dbi 2
ac  adi  bci  db
 ac  db    ad  bc  i
The formulas are scarier than it really is.
multiply z1z2
Example
Rectangular form
z1 z2
2

3  2i 3 2  3 2i
Where z1  2 3  2i  4  cos30  i sin 30 
z2  3 2  3 2i  6  cos 45  i sin 45
Trig form
z1 z2

6 6  6 6i  6 2i  6 2i 2
 4  cos 30  i sin 30   6  cos 45  i sin 45 
4  6  cos  30  45  i sin  30  45 
6 6  6 6i  6 2i  6 2
24  cos 75  i sin 75
6
r
 

6 6 2  6 6 6 2 i
6
6 6 2
  6
2
6 6 2

2
r  216  72 12  72  216  72 12  72
r  576  24
6 6 6 2 
  75
6 6 6 2 
  tan 1 
Dividing Complex Numbers
In rectangular form, you rationalize
using the complex conjugate.
a  bi
c  di
 a  bi   c  di 



 c  di   c  di 
ac  adi  bci  bdi 2
c 2  d 2i 2
ac  adi  bci  bd
c2  d 2
ac  bd  bc  ad 
 2
i
2
2
2 
c d
 c d 
In trig form, you just divide the rs and
subtract the theta.
r1  cos 1  i sin 1 
r2  cos  2  i sin  2 
r1
cos 1   2   i sin 1   2  

r2
divide
Where z1  3 2  3 2i  6  cos 45  i sin 45
Example
z2  2 3  2i  4  cos 30  i sin 30 
Rectangular form
Trig form
3 2  3 2i
2 3  2i
6  cos 45  i sin 45
4  cos 30  i sin 30 
 3 2  3 2i   2 3  2i 



2
3

2
i
2
3

2
i



6 6  6 2i  6 6i  6 2i 2
12  4i 2
6 6  6 2i  6 6i  6 2
12  4
6
6 6 2
16
  6
6
cos  45  30   i sin  45  30  

4
3
 cos15  i sin15
2

6 6 2 i
16
z1
z2




 6 6 6 2 


1 
16
  15
  tan
 6 6 6 2 


216  72 12  72  216  72 12  72
16


r


2


 6 6 6 2   6 6 6 2 
 

r 
16
16

 


 

256
r
576
9 3


256
4 2
2
De Moivre’s Theorem



If z  r (cos   i sin  ) is a complex
number
And n is a positive integer
n
n
Then z  r (cos   i sin  )


 r (cos n  i sin n )
n
Who was De Moivre?
A brilliant French mathematician who
was persecuted in France because of
his religious beliefs. De Moivre moved to
England where he tutored mathematics
privately and became friends with Sir
Issac Newton.
De Moivre made a breakthrough in the field of probability
(writing the Doctrine of Chance), but more importantly he
moved trigonometry into the field of analysis through
complex numbers with De Moivre’s theorem.
But, can we prove DeMoivre’s Theorem?
Let’s look at some Powers of z.
z  r(cos  i sin  )
z   r (cos   i sin  ) 
2
2
 r 2 (cos  i sin  )2
 r 2 (cos  i sin  )(cos  i sin  )
 r 2 (cos2   2i cos sin   i 2 sin2  )
 r 2 (cos2   sin2   2i cos sin  )
 r (cos2  i sin 2 )
2
Let’s look at some more Powers of z.
z   r (cos   i sin  ) 
3
3
 r(cos  i sin )r 2 (cos  i sin  )2
 r (cos3  i sin 3 )
3
z   r (cos   i sin  ) 
4
4
 r(cos  i sin  )r3 (cos  i sin  )3
 r 4 (cos4  i sin 4 )
It appears that:
cos  i sin  cosn  i sinn
n
Proof:
 n=1, then the statement is true.
Assume
We can continue in the previous manor up to some arbitrary k
cos  i sin  cosk  i sink
cos  i sin
k
Let n = k, so that:
Now find
k1
cos  i sin  cosk  i sinkcos  i sin

cos


i sin  cosk cos  sink sin icosk sin  sink cos

k1
k1


cos  i sin
cos  i sin
k1
 cos(k  )  i sin(k  )
k1
 cos(k  1)  i sin(k  1)
Euler’s Formula e  cos  i sin 
i
We can also use Euler’s formula to prove DeMoivre’s Theorem.
 cos  i sin  
n
 e
=e

i n
in
=  cos n  i sin n 
So what is the use?
Find an identity for cos5 using Mr. De Moivre’s fantastic theory

cos5  i sin5  cos  i sin

5
Remember the binomial expansion:

(a  b)5  1(a)5 (b)0  5(a)4 (b)1  10(a)3 (b)2  10(a)2 (b)3  5(a)1(b)4  1(a)0 (b)5
Apply it:
cos5  i sin 5  (1)(cos5  )  (5)(cos4  )(i sin  )  (10)(cos3  )(i 2 sin 2  )  (10)(cos2  )(i 3 sin 3  )
(5)(cos)(i4 sin4 )  (1)(i5 sin5 )
Cancel out the imaginery numbers:

cos5  cos5   10cos3  sin2   5 cos sin4 
Now try these:
cos3
 cos   3cos  sin 
sin3
 3cos  sin  sin 
3
2
2
3

sin4

 4 cos  sin  4 cos  sin 
3
3
Powers of Complex Numbers
This is horrible in rectangular
form.
 a  bi 
 a  bi  a  bi  a  bi  ...  a  bi 
n
The best way to expand one
of these is using Pascal’s
triangle and binomial
expansion.
You’d need to use an i-chart
to simplify.
It’s much nicer in trig form. You just
raise the r to the power and multiply
theta by the exponent.
z  r  cos   i sin  
z n  r n  cos n  i sin n 
Example
z  5  cos 20  i sin 20 
z 3  53  cos3  20  i sin 3  20 
z 3  125  cos60  i sin 60 
Roots of Complex Numbers

There will be as many answers as the
index of the root you are looking for



Square root = 2 answers
Cube root = 3 answers, etc.
Answers will be spaced symmetrically
around the circle

You divide a full circle by the number of
answers to find out how far apart they are
The formula
z  r  cos   i sin  
n
Using DeMoivre’s Theorem we get
  360k
  360k 

z  n r  cos
 i sin
 or
n
n


k starts at 0 and goes up to n-1
This is easier than it looks.
n
  2 k
  2 k 

r  cos
 i sin

n
n


General Process
1.
2.
3.
4.
Problem must be in trig form
Take the nth root of r. All answers
have the same value for r.
Divide theta by n to find the first
angle.
Divide a full circle by n to find out how
much you add to theta to get to each
subsequent answer.
Example
Find the 4th root of z  81cos80  i sin80
1. Find the 4th root of 81
r  4 81  3
2. Divide theta by 4 to get the
first angle.

3. Divide a full circle (360) by
4 to find out how far apart the
answers are.
360
 90 between answers
4
4.
List the 4 answers.
•
•
80
 20
4
z1  3  cos 20  i sin 20 
The only thing that changes z2  3  cos  20  90   i sin  20  90    3  cos110  i sin110 
is the angle.
z3  3  cos 110  90  i sin 110  90    3  cos 200  i sin 200 
The number of answers
z  3  cos  200  90   i sin  200  90    3  cos 290  i sin 290 
equals the number of roots. 4