6.6 DeMoivre’s Theorem
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Transcript of 6.6 DeMoivre’s Theorem
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6.6 DeMoivre’s Theorem
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I. Trigonometric Form of Complex Numbers
A.) The standard form of the complex number
is very similar to the component form of a vector
If we look at the trigonometric form of v, we can see
.ai bj v
z a b i
cos sini j v v
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P (a, b)
B.) If we graph the complex z = a + bi on the complex plane, we can see the similarities with the polar plane.
z = a + bi
θ
r
a
b
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C.) If we let and then,
where
sin b r cosa r
2 2 ,r z a b
cos sinz a b r r i i
tan and 1b
a i
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D.) Def. – The trigonometric form of a complex number z is given by
Where r is the MODULUS of z and θ is the ARGUMENT of z.
cos sin
or
z r
z rcis
i
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E.) Ex.1 - Find the trig form of the following:1 3
2.) 2 2
i1.) 2i
2 2
1
0 2 2
2tan
0 2
0 2
(cos sin )
2 cos sin2 2
22
r
z i
r
cis
i
i
22
1
1 31
2 2
5tan 3
3
1 3
2 2(cos sin )
5 51 cos sin
3 3
5
3
r
z i
r
cis
i
i
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A.) Let .
Mult.-
Div. -
1 1 1 1 2 2 2 2cos sin and cos sinz r i z r i
1 2 1 2 1 2 1 2cos sinz z r r i
II. Products and Quotients
1 11 2 1 2
2 2
cos sinz r
iz r
DERIVE THESE!!!!
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B.) Ex. 2 – Given .
find 11 2
2
and z
z zz
1 2 5 2 cos 120 sin 120
1 310
2 2
5 5 3
z z i
i
i
1
2
5cos 90 sin 90
2
50
25
2
zi
z
i
i
1
2
5 cos15 sin15 &
2 cos105 sin105
z i
z i
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III. Powers of Complex Numbers
A.) DeMoivre’s (di-’moi-vərz) Theorem –
If z = r(cosθ + i sinθ) and n is a positive integer, then,
(cos sin ) cos sinnn nz r i r n i n
Why??? – Let’s look at z2-
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2z z z
2 cos sin cos sinz r i r i
2 2 cos sin cos sinz r i i
2 2 2 2 2cos 2 cos sin sinz r i i
2 2 cos 2 sin 2z r i
2 2 2 2cos 1 sin 2cos sinz r i
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B.) Ex. 3 – Find by “Foiling” 3
1 3i
1 3 1 3 1 3i i i
1 2 3 3 1 3i i
2 2 3 1 3i i
8
2 6
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C.) Ex. 4– Now find using DeMoivre’s Theorem
3
1 3i
2 3
r
3
3 2 cos sin3 3
z i
3 32 cos3 sin 33 3
z i
3 8( 1 0) 8z
3 8 cos sinz i
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D.) Ex. 5 –Use DeMoivre’s Theorem to simplify10
3 3
2 2i
2
3 62 =
2 2 4r
10
10 6 10 10cos sin
2 4 4z i
510
10
6 5 5cos sin
2 2 2z i
510
10
6
2z i
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IV. nth Roots of Complex Numbers
A.) Roots of Complex Numbers –
v = a + bi is an nth root of z iff vn = z .
If z = 1, then v is an nth ROOT OF UNITY.
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B.) If , then the n distinct complex numbers
cos sinz r i
+ 2 + 2cos sinn k k
rn n
i
Where k = 0, 1, 2, …, n-1 are the nth roots of the complex number z.
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C.) Ex. 6- Find the 4th roots of
41
2164
z cis
16 .2
z cis
4
3
4216
4z cis
42
2216
4z cis
44
6216
4z cis
1 28
z cis
3
92
8z cis
2
52
8z cis
4
132
8z cis
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A.) Ex. 7 - Find the cube roots of -1.
2( 1) 1 0z z z
3 1z
V. Finding Cube Roots
3 1 0z 1 1 4(1)(1)1 or
2(1)z z
1 3 1 31, ,
2 2 2 2z i i
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Now....
1
1 3cos sin
3 3 2 2z i
i
1 0z i
1 cos sinz i
2
3 3cos sin 1 0
3 3z i i
3
5 5 1 3cos sin
3 3 2 2z i
i
Plot these points on the complex plane. What do you notice about them?
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1
-1
2
1z
2z
3z
Equidistant from the origin and equally spaced about the origin.
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VI. Roots of Unity
A.) Any complex root of the number 1 is also known as a ROOT OF UNITY.
B.) Ex. 8 - Find the 6 roots of unity.
1 0 0z i cis
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1
0
6z cis
2
0 2
6z cis
3
0 4
6z cis
1
3cis
1 3
2 2i
2
3cis
1 3
2 2i
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4
0 6
6z cis
6
0 10
6z cis
5
0 8
6z cis
cis 1
4
3cis
1 3
2 2i
5
3cis
1 3
2 2i