6(cos 300 i sin 300 ) 4(cos 240 i sin 240 ) 24(cos 540 i sin 540 ) i sin 300 ) i sin 150 ) 2(cos 150 in .NET framework

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1 2 i 23
2(cos 180 cos 30 6(cos 90 22(cos 225
3 22(cos 225 1 4A i
i sin 225 ) i sin 120 )
i23) (1
i 23)
2(cos 60 4(cos 120
i sin 60 ) 2(cos 60
2i 23
CHAPTER 15 Complex Numbers
15.17 Verify De Moivre s theorem for n
Let z For n For n r(cos 2: z2 r2[(cos 3: z
2 and n
i sin ). [r(cos sin ) z
i sin )][r(cos i(2 sin cos )] sin 2 sin ) [r (cos 2
i sin )] r2(cos 2 i sin 2 ) i sin )] cos 2 sin )]
i sin 2 )][r(cos i(sin 2 cos
r3[(cos 2 cos r (cos 3
i sin 3 )
The theorem may be established for n a positive integer by mathematical induction.
15.18 Evaluate each of the following using De Moivre s theorem and express each result in rectangular form: (a) (1
(a) (1 (b) ( !3 (c) ( 1
i 23)4,
i23)4 i)5 i)
10 4
(b) ( 23
[2(cos 60 2 (cos 240
i)5, (c) ( 1
i sin 60 )]4 i sin 240 ) 8
i)10
i sin 4 60 ) i sin 1650 ) i sin 270 ) 32 32i 16 !3 16i 8i 23 32(cos 270
24(cos 4 60 32(cos 1650
[2(cos 330 [ 22(cos 135
i sin 330 )]5
i sin 135 )]10
15.19 Find the indicated roots in rectangular form, except when this would necessitate the use of tables or a calculator. (a) (b) (c) (d)
(a) and Putting k (2
Square roots of 2 2i 23 Fourth roots of 8 8i23 Cube roots of 422 4i22 Cube roots of 1
2 2i 23 2i 23)1>2 R1 R2
(e) Fourth roots of i (f) Sixth roots of 1 (g) Fourth roots of 16i
4[cos (300 2[cos (150 k 360 ) k 180 ) 2A 2A
i sin (300 i sin (150 23
1 2i 1 2i
k 360 )] k 180 )]
0 and 1, the required roots are 2(cos 150 2(cos 330 8i 23 8i23)1>4 R1 R2 R3 R4 2(cos 60 2(cos 150 2(cos 240 2(cos 330 4i 22 4i 22)1>3 R1 R2 R3 2(cos 45 2(cos 165 2(cos 285 i sin 150 ) i sin 330 )
23 23 i
1 2 23
(b) and Putting k ( 8
16[cos (240 2[cos (60 i sin 60 ) i sin 150 ) i sin 240 ) i sin 330 )
k 360 ) k 90 ) 2 A1 2 2A 2A
1 2 1 2 1 2
i sin (240
k 360 )]
i sin (60 i1 23 B 2 23
1 2i
k 90 )] 1
1 2i
0, 1, 2, and 3, the required roots are i 23 23 1 23 k 120 )] 22 i 22 i 23 i i
i1 23 2
2 A 23
(c) and Putting k
422 ( 422
8[cos (135 2[cos (45 i sin 45 )
k 360 ) k 120 )
i sin (135 i sin (45
k 360 )]
0, 1, and 2, the required roots are 2 A 1> 22 i> 22 B i sin 165 ) i sin 285 ) cos (k 120 ) i sin (k 120 ).
(d) 1
cos (0
k 360 )
i sin (0 R1 R2 R3
k 360 ) and 11/3 i sin 0 1
Putting k
0, 1, and 2, the required roots are cos 0 cos 120 cos 240 i sin 120 i sin 240
1 2 1 2
i 1 23 2 i 1 23 2
CHAPTER 15 Complex Numbers
Note that and (e) i
R2 2 R2 3 R2R3
cos 2(120 ) cos 2(240 ) (cos 120 k 360 ) cos 22 1 2 cos 112 1 2
i sin 2(120 ) i sin 2(240 )
3 R3
R2 i sin 240 ) cos 221 2 cos 202 1 2 cos 292 1 2 cos 0 k 90 i sin 0 R1 k 90 . i sin 221 2
i sin 120 )(cos 240 i sin (90 i sin 22 1 2 i sin 112 1 2
cos (90
k 360 ) and i1/4
Thus, the required roots are R1 R2 (f) R3 R4 i sin 202 1 2 i sin 292 1 2
1 cos (180 k 360 ) i sin (180 k 360 ) k 60 ) i sin (30 k 60 ). and ( 1)1/6 cos (30 Thus, the required roots are R1 R2 R3 R4 R5 R6 Note that R2 2 R2 5 cos 180 cos 30 cos 90 cos 150 cos 210 cos 270 cos 330 i sin 30 i sin 90 i sin 150 i sin 210 i sin 270 i sin 330
1 2 23 1 2i
1 2 23 1 2 23 1 2i 1 2i
1 2 23 1 2i
i sin 180 and thus R2 and R5 are the square roots of
1; that i.
R3 1 R3 2 (g) and
R3 3 R3 4
R3 5 R3 6
cos 90 cos 270
i sin 90 i sin 270
i and thus R1, R3, and R5 are the cube roots of i; and that i and thus R2, R4, and R6 are the cube roots of
16i ( 16i)1/4
16[cos (270 k 360 ) i sin (270 k 360 )] 1 1 2 cos 672 k 90 i sin 672 k 90
1 1 1 1 1 1
Thus, the required roots are R1 R2 2 cos 672 2 cos 1572 i sin 672 i sin 1572 R3 R4 2 cos 2472 2 cos 3372 i sin 2472 i sin 3372
SUPPLEMENTARY PROBLEMS
15.20 Perform the indicated operations, writing the results in the form a (a) (6 (b) (6 (c) (3 (d) (3 (e) 3(2 (f) 2i(3 (g) (2 (h) (2 (i) (3 (j) (2 (k) (2 (l) (1 2i) 2i) 2i) 2i) i) 4i) 3i)(1 3i)(5 2i)( 4 3i)(3 (2 (2 ( 4 (4 6 3i) 3i) 3i) 3i) 3i 8 2i) 2i) i) 2i) 12 6i 4 16 10 5i (6 8 4 25) 3 23i (3 25 8) i 7i 11i 11i (r) (q) 1 8 4 i 5i 1 i (p) i (m) (2 (n) (4 (o) (1 2 1 3 3 3 2 i)
3 12 3i) 5 2 17 25
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