6.6
De Moivre’s
Theorem and
nth Roots
Copyright © 2011 Pearson, Inc.
What you’ll learn about





The Complex Plane
Trigonometric Form of Complex Numbers
Multiplication and Division of Complex Numbers
Powers of Complex Numbers
Roots of Complex Numbers
… and why
The material extends your equation-solving technique
to include equations of the form zn = c, n is an integer
and c is a complex number.
Copyright © 2011 Pearson, Inc.
Slide 6.6 - 2
Complex Plane
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Slide 6.6 - 3
Absolute Value (Modulus) of a
Complex Number
The absolute value or modulus of a complex number
z = a + bi is | z |=| a + bi |= a + b .
2
2
In the complex plane, | a + bi | is the distance of a + bi
from the origin.
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Slide 6.6 - 4
Graph of z = a + bi
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Slide 6.6 - 5
Trigonometric Form of a Complex
Number
The trigonometric form of the complex number
z = a + bi is
(
z = r cosq + isin q
)
where a = r cosq , b = r sin q , r = a 2 + b2 ,
and tan q = b / a. The number r is the absolute
value or modulus of z, and q is an argument of z.
Copyright © 2011 Pearson, Inc.
Slide 6.6 - 6
Example Finding Trigonometric
Form
Find the trigonometric form with 0 £ q < 2p for the
complex number 1+ 3i.
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Slide 6.6 - 7
Example Finding Trigonometric
Form
Find the trigonometric form with 0 £ q < 2p for the
complex number 1+ 3i.
Find r: r =|1+ 3i |= 12 +
( 3)
2
= 2.
3
p
Find q : tan q =
so q = .
1
3
æ
p
pö
Therefore, 1+ 3i = 2 ç cos + isin ÷ .
3
3ø
è
Copyright © 2011 Pearson, Inc.
Slide 6.6 - 8
Product and Quotient of Complex
Numbers
(
)
(
)
Let z1 = r1 cosq1 + isin q1 and z2 = r2 cosq 2 + isin q 2 .
Then
(
)
(
)
1. z1 × z2 = r1r2 éëcos q1 + q 2 + isin q1 + q 2 ùû .
z1 r1
2.
= éëcos q1 - q 2 + isin q1 - q 2 ùû , r2 ¹ 0.
z2 r2
(
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)
(
)
Slide 6.6 - 9
Example Multiplying Complex
Numbers
Express the product of z1 and z2 in standard form.
æ
æ
p
pö
p
pö
z1 = 4 ç cos + isin ÷ , z2 = 2 ç cos + isin ÷
4
4ø
6
6ø
è
è
Copyright © 2011 Pearson, Inc.
Slide 6.6 - 10
Example Multiplying Complex
Numbers
Express the product of z1 and z2 in standard form.
æ
æ
p
pö
p
pö
z1 = 4 ç cos + isin ÷ , z2 = 2 ç cos + isin ÷
4
4ø
6
6ø
è
è
z1 × z2 = r1r2 éëcos q1 + q 2 + isin q1 + q 2 ùû
(
)
(
)
é æp pö
æp pöù
= 4 2 êcos ç + ÷ + isin ç + ÷ ú
è 4 6øû
ë è 4 6ø
é æ 5p ö
æ 5p ö ù
= 4 2 êcos ç ÷ + isin ç ÷ ú
è 12 ø û
ë è 12 ø
(
)
» 4 2 0.259 + i0.966 » 1.464 + 5.464i
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Slide 6.6 - 11
A Geometric Interpretation of z2
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Slide 6.6 - 12
De Moivre’s Theorem
(
)
Let z = r cosq + isin q and let n be a positive integer.
Then
(
)
n
(
)
z = éë r cosq + isin q ùû = r n cos nq + isin nq .
n
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Slide 6.6 - 13
Example Using De Moivre’s Theorem
4
æ
3 1ö
Find ç - i ÷ using De Moivre's theorem.
2ø
è 2
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Slide 6.6 - 14
Example Using De Moivre’s Theorem
4
æ
3 1ö
Find ç - i ÷ using De Moivre's theorem.
2ø
è 2
3 1
7p
The argument of z = - i is q =
,
2
2
6
3 1
and its modulus -i =
2
2
3 1
+ = 1.
4 4
Hence,
7p
7p
z = 2cos
+ isin
6
6
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Slide 6.6 - 15
Example Using De Moivre’s Theorem
4
æ
3 1ö
Find ç - i ÷ using De Moivre's theorem.
2ø
è 2
æ 7p ö
æ 7p ö
4
z = cos ç 4 × ÷ + isin ç 4 × ÷
6 ø
6 ø
è
è
æ 14p ö
æ 14p ö
= cos ç
+ isin ç
÷
è 3 ø
è 3 ÷ø
æ 2p ö
æ 2p ö
= cos ç ÷ + isin ç ÷
è 3ø
è 3ø
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1
3
=- +i
2
2
Slide 6.6 - 16
nth Root of a Complex Number
A complex number v = a + bi is an nth root of z if
v n = z.
If z = 1, then v is an nth root of unity.
Copyright © 2011 Pearson, Inc.
Slide 6.6 - 17
Finding nth Roots of a Complex
Number
(
)
If z = r cosq + isin q , then the n distinct
complex numbers
n
æ
q + 2p k
q + 2p k ö
r ç cos
+ isin
,
÷
n
n ø
è
where k = 0,1,2,.., n - 1,
are the nth roots of the complex number z.
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Slide 6.6 - 18
Example Finding Cube Roots
Find the cube roots of 1.
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Slide 6.6 - 19
Example Finding Cube Roots
Find the cube roots of 1.
Write 1 in complex form: z = 1+ 0i = cos0 + isin0
The third roots of 1 are the complex numbers
0 + 2p k
0 + 2p k
cos
+ isin
for k = 0,1,2.
3
3
z1 = cos0 + isin0 = 1
2p
2p
1
3
z2 = cos
+ isin
=- +
i
3
3
2 2
4p
4p
1
3
z3 = cos
+ isin
=- i
3
3
2 2
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Slide 6.6 - 20
Quick Review
1. Write the roots of the equation x 2 + 12 = 6x in a + bi form.
(
)
3
2. Write the complex number 1+ i in standard form a + bi.
3. Find all real solutions to x 3 - 27 = 0.
Find an angle q in 0 £ q < 2p which satisfies both equations.
1
3
4. sin q = and cosq = 2
2
2
2
5. sin q = and cosq = 2
2
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Slide 6.6 - 21
Quick Review Solutions
1. Write the roots of the equation x + 12 = 6x in a + bi form.
2
3 + 3i, 3 - 3i
(
)
3
2. Write the complex number 1+ i in standard form a + bi.
-2 + 2i
3. Find all real solutions to x 3 - 27 = 0. x = 3
Find an angle q in 0 £ q < 2p which satisfies both equations.
1
3
4. sin q = and cosq = 2
2
2
2
5. sin q = and cosq = 2
2
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q = 5p / 6
q = 5p / 4
Slide 6.6 - 22
Chapter Test
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Slide 6.6 - 23
Chapter Test
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Slide 6.6 - 24
Chapter Test
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Slide 6.6 - 25
Chapter Test Solutions
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Slide 6.6 - 26
Chapter Test Solutions
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Slide 6.6 - 27
Chapter Test
7. Find a parameterization for the line through the points
( - 1,-2) and (3,4).
x = 2t + 3, y = 3t + 4
5
é æ
p
pöù
8. Use De Moivre's theorem to evaluate ê3ç cos + isin ÷ ú .
4
4øû
ë è
Write your answer in (a) trigonometric form and (b) standard
form.
æ
5p
5p ö
-243 2 -243 2
(a) 243ç cos
+ isin ÷ (b)
+
i
4
4 ø
2
2
è
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Slide 6.6 - 28
Chapter Test
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Slide 6.6 - 29