Download Math 112 – Elementary Functions

Math 112
Elementary Functions
Chapter 7 – Applications of Trigonometry
Section 3
Complex Numbers: Trigonometric Form
Graphing Complex Numbers
How do you graph a real number?

Use a number line.

The point corresponding to a real number represents the
directed distance from 0.
y
y is a negative
real number
0
1
x
x is a positive
real number
Graphing Complex Numbers
General form of a complex number …
a + bi
a  R and
 i = -1

bR
Therefore, a complex number is essentially an ordered pair!
(a, b)
Graphing Complex Numbers
Imaginary
Axis
-4
2
Real Axis
All real numbers, a
= a+0i, lie on the
real axis at (a, 0).
Graphing Complex Numbers
Imaginary
Axis
All imaginary numbers, bi = 0+bi,
lie on the imaginary axis at (0, b).
2i
Real Axis
-4i
Graphing Complex Numbers
Imaginary
Axis
2 + 3i
-4 + i
Real Axis
3 – 2i
All other numbers,
a+bi, are located at
the point (a,b).
-3 - 4i
Absolute Value
Real Numbers:
|x| = distance from the origin
 x if x  0
x 
 x if x  0
x x
2
Absolute Value
Complex Numbers:
|a + bi| = distance from the origin
a + bi
b
a  bi  a  b
2
2
a
Note that if b = 0, then this reduces to an equivalent definition for the absolute value
of a real number.
Trigonometric Form of a
Complex Number
r  a  bi  a  b
2
a + bi
r
b

a
2
a
cos 
r

a  r cos 
b
sin  
r

b  r sin 
Therefore,
a + bi = r (cos + i sin)
Note: As a standard,  is to be the smallest positive number possible.
Trigonometric Form of a
Complex Number
a  bi  r (cos   i sin  )
 r cis
Steps for finding the trig
form of a + bi.
• r = |a + bi|
•  is determined by …
cos  = a / r
sin  = b / r
Example:
2 – 3i
r  22   3  13
2
  cos
  sin
1
1
2
 .98  56
13
3
 .98  56
13
Therefore, 2  3i  13 cis 5.3  13 cis 304
Trigonometric Form of a
Complex Number – Determining 
a + bi = r cis 
r = |a+bi|
cos  = a/r
Using cos  = a/r
sin  = b/r
Using sin  = b/r
• Q1:  = cos-1(a/r)
• Q1:  = sin-1(b/r)
• Q2:  = cos-1(a/r)
• Q2:  = 180° - sin-1(b/r)
• Q3:  = 360° - cos-1(a/r)
• Q3:  = 180° - sin-1(b/r)
• Q4:  = 360° - cos-1(a/r)
• Q4:  = 360° + sin-1(b/r)
For Radians, replace 180° with  and 360° with 2.
Trigonometric Form of Real and
Imaginary Numbers (examples)
Real/Imaginary
Number
Complex
Form
Trig w/
Degrees
Trig w/
Radians
0
0 + 0i
0 cis 0°
0 cis 0
2
2 + 0i
2 cis 0°
2 cis 0
-5
-5 + 0i
5 cis 180°
2 cis 
3i
0 + 3i
3 cis 90°
3 cis (/2)
-4i
0 – 4i
4 cis 270°
4 cis (3/2)
Converting the Trigonometric Form
to Standard Form
r cis 
= r (cos  + i sin )
= (r cos ) + (r sin ) i
Example: 4 cis 30º
= (4 cos 30º) + (4 sin 30º)i
= 4(3)/2 + 4(1/2)i
= 23 + 2i
 3.46 + 2i
Arithmetic with Complex Numbers
Addition & Subtraction

Standard form is very easy ………Trig. form is ugly!
Multiplication & Division

Standard form is ugly…………….Trig. form is easy!
Exponentiation & Roots

Standard form is very ugly….Trig. form is very easy!
Multiplication of Complex Numbers
(Standard Form)
a  bi c  di   ac  bd   ad  bc i
Multiplication of Complex Numbers
(Trigonometric Form)
r cos  i sin    s(cos   i sin  )
 rscos cos   sin  sin   i sin  cos   i cos sin  
 rscos     i sin    
r cis  s cis    rs cis    
Division of Complex Numbers
(Standard Form)
a  bi (ac  bd )  (bc  ad )i

2
2
c  di
c d
Division of Complex Numbers
(Trigonometric Form)
r (cos   i sin  )
s(cos   i sin  )

r (cos   i sin  ) (cos   i sin  )

s(cos   i sin  ) (cos   i sin  )
r  cos  cos   sin  sin   i sin  cos   i cos  sin  
 

s
(cos 2   sin 2  )


r
cos     i sin    
s
r cis  r
 cis    
s cis  s
Powers of Complex Numbers
(Trigonometric Form)
[r cis ]2
[r cis ]3
= (r cis ) • (r cis )
= (r cis )2 • (r cis )
= r2 cis( + )
= r2 cis(2) •(r cis )
= r2 cis 2
= r3 cis 3
Powers of Complex Numbers
(Trigonometric Form)
DeMoivre’s Theorem
(r cis
n
)
=
n
r
cis (n)
Roots of Complex Numbers
An nth root of a number (a+bi) is any solution to
the equation …
xn = a+bi
Roots of Complex Numbers
Examples

The two 2nd roots of 9 are …


32 = 9
and
(-3)2 = 9
and
(-5i)2 = -25
The two 2nd roots of -25 are …


3 and -3, because:
5i and -5i, because:
(5i)2 = -25
The two 2nd roots of 16i are …

22 + 22i and -22 - 22i
because (22 + 22i)2 = 16i and
(-22 - 22i)2 = 16i
Roots of Complex Numbers
Example:
Find all of the
 x4
th
4
roots of 16.
= 16
 x4 – 16 = 0
 (x2 + 4)(x2 – 4) = 0
 (x + 2i)(x – 2i)(x + 2)(x – 2) = 0
 x = ±2i or ±2
Roots of Complex Numbers
In general, there are always …
n “nth roots” of any complex number
Roots of Complex Numbers
One more example …
3

8 cis 75
Using
DeMoivre’s
Theorem
Let k = 0, 1, & 2

 8 cis 75


1
3


 8 cis 75  360 k
 75  360 k 

 2 cis 
3






3
 2 cis 25  120 k
 2 cis 25 , 2 cis 145 , 2 cis 265

1

NOTE: If you let k = 3, you get 2cis385 which is equivalent to 2cis25.


Roots of Complex Numbers
The n nth roots of the complex number r(cos  + i sin ) are …
 

360 
360 
  i sin   k 

r cos  k 
n 
n 
n
 n
1
n
where k  0, 1, 2, 3, 4, ..., n  1
Roots of Complex Numbers
The n nth roots of the complex number r cis  are …
   360 k 

r cis 
n


1
n
or
   2k 
r cis 

 n 
1
n
where k  0, 1, 2, 3, 4, ..., n  1
Summary of (r cis ) w/ r = 1
cis Acis B  cis A  B
cis A
 cis  A  B 
cis B
cis A
n
 cis nA
Does this
remind you
of something?
x a x b  x a b
xa
a b

x
xb
x 
a b
 x ab
Euler’s Formula
i
e  cos  i sin 
Note:  must be expressed in radians.
Therefore, the complex number …
a  bi  r (cos  i sin  )
r = |a + bi|
 r cis 
cos  = a/r
 re i
sin  = b/r
Results of Euler’s Formula
i
e  1

This gives a relationship
between the 4 most
common constants in
mathematics!
i
e 1  0
Results of Euler’s Formula
e
i

2
 i



ii is a real number!

i
 i2 
 e   ii
 
 
e


2
i
i
i  0.2078795763...
i
Results of Euler’s Formula
e e
2
 ix
e e
2
 ix
ix
ix

cos x

sin x