Find all fifth roots of 4 in .NET framework

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EXAMPLE 15.9 Find all fifth roots of 4
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4i. i sin 315 ), but we shall need the more general form k360 ) i sin (315 k360 )]
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The usual polar form of 4
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4i is 4 22(cos 315 4 22 [cos (315
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where k is any integer, including zero. Using De Moivre s theorem, a fifth root of 4 {4 22[cos (315 k360 )
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4i is given by k360 )]}1/5 k 360 5 k72 ) i sin 315 5 k72 )] k360 b
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i sin (315 315
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A 4 22 B 1>5 a cos
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22 [ cos (63 Assigning in turn the values k 0, 1, 2, . . . , we find k k k k k k 0: 22(cos 63 1: 22(cos 135 2: 22(cos 207 3: 22(cos 279 4: 22(cos 351
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i sin (63
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i sin 63 ) i sin 135 ) i sin 207 ) i sin 279 ) i sin 351 ) i sin 423 )
R1 R2 R3 R4 R5
5: 22(cos 423 22(cos 63
i sin 63 )
R1, etc. 1) to k.
Thus, the five fifth roots are obtained by assigning the values 0, 1, 2, 3, 4 (i.e., 0, 1, 2, 3, . . . , n
(See also Prob. 15.19.)
The modulus of each of the roots is 22; hence these roots lie on a circle of radius 22 with center at the origin. The difference in amplitude of two consecutive roots is 728; hence the roots are equally spaced on this circle, as shown in Fig. 15.4.
Fig. 15.4
CHAPTER 15 Complex Numbers
SOLVED PROBLEMS
In Probs. 15.1 to 15.6, perform the indicated operations, simplify, and write the result in the form a 15.1 (3 15.2 (4 15.3 (2 15.4 (3 15.5 15.6 1 2 3 2 4i) 2i) i)(3 4i)(3 3i i 2i 3i ( 5 ( 1 2i) 4i) (1 (2 (3 (2 7i) 3i) (6 9 (3 [4 2) 16 5) ( 4 (2 3)i 8 7)i 3)i i 2 5 3i i
( 1)] ( 4 25 i) i) 3 i) 3i) yi 4 (2
3i)(2 i)(2 2i)(2 3i)(2
3) 4 (6 3i.
( 1 1 6) 4 (9 9
6)i 4)i
1 12 13
i 5 i 13
15.7 Find x and y such that 2x
Here 2x 4 and y
3; then x
2 and y
15.8 Show that the conjugate complex numbers 2 4x 5 0.
i and 2
i are roots of the quadratic equation x2
0. 0.
For x 2 i: (2 i)2 4(2 i) 5 4 4i i2 8 4i 5 For x 2 i: (2 i)2 4(2 i) 5 4 4i i2 8 4i 5 Since each number satisfies the equation, it is a root of the equation.
15.9 Show that the conjugate of the sum of two complex numbers is equal to the sum of their conjugates.
Let the complex numbers be a sum is (a c) (b d)i. bi and c di. Their sum is (a bi and c d)i (a c) (b d)i and the conjugate of the
The conjugates of the two given numbers are a (a c) ( b
di, and their sum is c) (b d)i
15.10 Represent graphically (as a vector) the following complex numbers: (a) 3 2i, (b) 2 i, (c) 2 i, (d) 1 3i
1), ( 2, 1), ( 1, 3) and join each to the We locate, in turn, the points whose coordinates are (3, 2), (2, origin O.
15.11 Perform graphically the following operations: (a) (3 4i) (2 5i), (b) (3 4i) (2 3i), (c) (4 3i) (2 i), (d) (4 3i) (2 i)
For (a) and (b), draw as in Fig. 15.5(a) and (b) the two vectors and apply the parallelogram law. For (c), draw the vectors representing 4 3i and 2 i and apply the parallelogram law as in Fig. 15.5(c). For (d), draw the vectors representing 4 3i and 2 i and apply the parallelogram law as in Fig. 15.5(d).
15.12 Express each of the following complex numbers z in polar form: (a) 1 i 23, (b) 6 23 6i, (c) 2
#( 1)2
2i, (d) 3
( 23)2
0i, (e) 4i
23>( 1)
4i, (f)
23 and u
(a) P lies in the second quadrant; r Thus, z 2(cos 120 i sin 120 ).
2; tan u 6/6 23
(b) P lies in the first quadrant; r #(6 23)2 Thus, z 12(cos 30 2 22(cos 315 i sin 30 ). 222
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