Transcript Algebra and Trigonometry - North Carolina Central University

Complex Numbers in Polar Form;
DeMoivre’s Theorem
Dr .Hayk Melikyan
Departmen of Mathematics and CS
[email protected]
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The Complex Plane
We know that a real number can be represented as a point on a number line.
By contrast, a complex number z = a + bi is represented as a point (a, b) in a
coordinate plane, shown below. The horizontal axis of the coordinate plane is
called the real axis. The vertical axis is called the imaginary axis. The
coordinate system is called the complex plane. Every complex number
corresponds to a point in the complex plane and every point in the complex
plane corresponds to a complex number.
Imaginary axis
b
z = a + bi
Real axis
a
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Text Example
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
c. z = -3
d. z = -4i
Solution
•
•
We plot the complex number z = 3 + 4i
the same way we plot (3, 4) in the
rectangular coordinate system. We
move three units to the right on the real
axis and four units up parallel to the
imaginary axis.
The complex number z = -1 – 2i
corresponds to the point (-1, -2) in the
rectangular coordinate system. Plot the
complex number by moving one unit
to the left on the real axis and two units
down parallel to the imaginary axis.
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5
4
3
2
1
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
1 2 3 4 5
-3
-4
-5
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Text Example cont.
Plot in the complex plane:
a. z = 3 + 4i
b. z = -1 – 2i
c. z = -3
d. z = -4i
Solution
• Because z = -3 = -3 + 0i, this complex
number corresponds to the point (-3, 0).
We plot –3 by moving three units to the
left on the real axis.
z = -3
•
Because z = -4i = 0 – 4i, this complex
number corresponds to the point (0, -4).
We plot the complex number by moving
three units down on the imaginary axis.
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5
4
3
2
1
-5 -4 -3 -2
z = -1 – 2i
z = 3 + 4i
1 2 3 4 5
-3
-4
-5
z = -4i
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The Absolute Value of a Complex Number
The absolute value of the complex number a + bi is
z  a  bi

a
2
 b
2
Determine the absolute value of z=2-4i
Solution:
z  a  bi 
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
2

20  2
2
 (4)
a
2

2
b
2
4  16
5
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Polar Form of a Complex Number
The complex number a + bi is written in polar form as
z = r (cos  + i sin  )
Where
a = r cos  ,
b = r sin  ,
2
r  a b
tan =b/a
2
The value of r is called the modulus (plural: moduli) of the
complex number z, and the angle  is called the argument of the
complex number z, with 0 <  < 2
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Text Example
Plot z = -2 – 2i in the complex plane. Then write z in polar form.
Solution
The complex number z = -2 – 2i, graphed below, is in rectangular form a + bi, with
a = -2 and b = -2. By definition, the polar form of z is r(cos  + i sin  ). We need to
determine the value for r and the value for  , included in the figure below.
Imaginary
axis
2
è
2
-2
Real
axis
r
-2
z = -2 – 2i
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Text Example cont.
Solution
r 
2
2
a b 
2
2
( 2)  ( 2 ) 
tan  
b
a

2
2
44 
8 2 2
1
Since tan 4 = 1, we know that  lies in quadrant III. Thus,
  

4

4
4


4

5
4
z  r (cos   i sin  )  2 2 (cos
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5
4
 i sin
5
4
)
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Product of Two Complex Numbers in Polar Form
Let z1 = r1 (cos 1+ i sin  1) and
z2 = r2 (cos  2 + i sin  2)
be two complex numbers in polar form. Their product, z1z2, is
z1z2 = r1 r2 (cos ( 1 +  2) + i sin ( 1 +  2))
To multiply two complex numbers, multiply moduli and add arguments.
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Text Example
Find the product of the complex numbers. Leave the answer in polar form.
z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)
Solution
z1z2 = [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]
Form the product of the given
numbers.
= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]
Multiply moduli and add
arguments.
= 28(cos 150º + i sin 150º)
Simplify.
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Quotient of Two Complex Numbers in Polar Form
Let z1 = r1 (cos 1 + i sin 1) and
z2 = r2 (cos 2 + i sin 2)
be two complex numbers in polar form.
Their quotient, z1/z2, is
z1

z2
r1
r2
[cos(  1   2 )  i sin(  1   2 )]
To divide two complex numbers, divide moduli and subtract
arguments.
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DeMoivre’s Theorem
•
Let z = r (cos  + i sin ) be a complex numbers in
polar form. If n is a positive integer, z to the nth
power, zn, is
z  [ r (cos   i sin  )]
n
n
 r (cos n   i sin n  )
n
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Text Example
Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi.
By DeMoivre’s Theorem,
Solution
[2 (cos 10º + i sin 10º)]6
=
26[cos
(6 · 10º) + i sin (6 · 10º)]
Raise the modulus to the 6th power and multiply
the argument by 6.
= 64(cos 60º + i sin 60º)
Simplify.
 1
3
 64 

i

2
 2
Write the answer in rectangular form.
 32  32
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3i




Multiply and express the answer in a + bi form.
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DeMoivre’s Theorem for Finding Complex Roots

Let =r(cos+isin) be a complex number in polar form. If
0,  has n distinct complex nth roots given by the formula
zk 
n
where
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
   360 k 
   360 k  
r  cos 
  i sin 

n
n





k  0 ,1, 2 , 3 ,..., n  1
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Example

Find all the complex fourth roots of
81(cos60º+isin60º)
zk 

4
n

   360 k 
r  cos 
  i sin
n



   360 k  


n



 60  360 * 0 
 60  360 * 0  
81  cos 
  i sin 

4
4





 3 (cos 15  i sin 15 )

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
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Objectives (sec 1.4 from math 1100 staff)
• Add and subtract complex numbers.
• Multiply complex numbers.
• Divide complex numbers.
• Perform operations with square roots of negative
numbers.
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Complex Numbers and Imaginary Numbers
The imaginary unit i is defined as
i
 1, w here i   1.
2
The set of all numbers in the form
a + bi
with real numbers a and b, and i, the imaginary unit,
is called the set of complex numbers.
The standard form of a complex number is
a + bi.
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Operations on Complex Numbers
The form of a complex number a + bi is like the binomial
a + bx. To add, subtract, and multiply complex numbers,
we use the same methods that we use for binomials.
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Example: Adding and Subtracting Complex Numbers
Perform the indicated operations, writing the result
in standard form:
a. (5  2 i )  (3  3 i )
 (5  3)  (  2  3)i
8i
b. (2  6 i )  (12  i )
 2  6 i  12  i
 (2  12)  (6  1)i
  10  7i
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Example: Multiplying Complex Numbers
Find the product:
a. 7 i (2  9 i )
 14 i  63 i
b. (5  4 i )(6  7 i )
2
 14 i  63(  1)
 14 i  63
 63  14i
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 5(6)  5(  7 i )  4 i (6)  4 i (  7 i )
 30  35 i  24 i  28 i
2
 30  11i  28(  1)
 30  11i  28
 58  11i
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Conjugate of a Complex Number
For the complex number a + bi, we define its complex
conjugate to be a – bi.
The product of a complex number and its conjugate
is a real number.
( a  bi )( a  bi )
 a ( a )  a (  bi )  bi ( a )  bi (  bi )
 a  abi  abi  b i
2
2 2
 a  b (  1)
2
2
a b
2
2
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Complex Number Division
The goal of complex number division is to obtain a real
number in the denominator. We multiply the numerator
and denominator of a complex number quotient by the
conjugate of the denominator to obtain this real number.
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Example: Using Complex Conjugates to Divide Complex Numbers
Divide and express the result in standard form:
5  4i
4i
5  4i 4  i

4i 4i
2
20  5 i  16 i  4 i

2
16  4 i  4 i  i


20  21i  (  4)
16  (  1)
16  21i
In standard form, the result is
16
17

21
i
17
17
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Principal Square Root of a Negative Number
For any positive real number b, the principal square root of
the negative number – b is defined by
b  i b
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Example: Operations Involving Square Roots of Negative
Numbers
Perform the indicated operations and write the result in
standard form:
a.
 27 
 48
 3i 3  4 i 3
 7i 3
b. (  2 
 3)
2
3 

   2  i 3   2  i 3 
 2 
2
 4  2 i 3  2 i 3  3i
2
 4  4 i 3  3(  1)
 1  4i 3
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