Solve for b if [b/(4+i√3)] - [b/(4-i√3)] = 1.?
回答 (8)
2018-02-06 5:40 pm
✔ 最佳答案
b [ 4 - i √3 ] - b [ 4 + i √3 ] = [ 4 - i√3 ] [ 4 + i √3 ]- 2 i √3 b = 19
b = - 19 / ( i 2√3 )
b = 19 i / (2√3)
b = 19√3 i / 6
2018-02-06 11:54 am
[b/(4+i√3)] - [b/(4-i√3)] = 1
Multiply by (4+i√3)(4-i√3)
b(4-i√3) -b(4+i√3) = 16 + 3
b(-2i√3) = 19
b = 19/(-2i√3)
b = -19i√3 /6
Multiply by (4+i√3)(4-i√3)
b(4-i√3) -b(4+i√3) = 16 + 3
b(-2i√3) = 19
b = 19/(-2i√3)
b = -19i√3 /6
2018-02-06 10:48 am
[b/(4 + i√3)] - [b/(4 - i√3)] = 1
[b(4 - i√3)] - [b(4 + i√3)] = [(4 + i√3)(4 - i√3)]
-2 i √3 b = 19
Complex solution:
b = (19 i)/(2√3)
[b(4 - i√3)] - [b(4 + i√3)] = [(4 + i√3)(4 - i√3)]
-2 i √3 b = 19
Complex solution:
b = (19 i)/(2√3)
2018-02-06 10:15 am
[b/(4+√-1√3)] - [b/(4-√-1√3] = 1
[b/(4+√-3)] - [b/(4-√-3] = 1
[b / (4 + 3i)] - [b / (4 - 3i)] = 1
[b / (4 + 3i)] = 1 + [b / (4 - 3i)] = (4 - 3i - b) / (4 - 3i)
4b - 3ib = 16 - 12i - 4b + 12i - 9i^2 - 3ib
4b = 16 - 4b +9
8b = 25
b = 3 1/8
[b/(4+√-3)] - [b/(4-√-3] = 1
[b / (4 + 3i)] - [b / (4 - 3i)] = 1
[b / (4 + 3i)] = 1 + [b / (4 - 3i)] = (4 - 3i - b) / (4 - 3i)
4b - 3ib = 16 - 12i - 4b + 12i - 9i^2 - 3ib
4b = 16 - 4b +9
8b = 25
b = 3 1/8
2018-02-06 10:08 am
[b / (4 + i√3)] - [b / (4 - i√3)] = 1
[b(4 - i√3) / (4 + i√3)(4 - i√3)] - [b(4 + i√3) / (4 + i√3)(4 - i√3)] = 1
[(4b - bi√3) / (16 + 3)] - [(4b + bi√3) / (16 + 3)] = 1
[(4b - bi√3) - (4b + bi√3)] / 19 = 1
(4b - bi√3 - 4b - bi√3) / 19 = 1
-2bi(√3) / 19 = 1
b = (-19 / 2i√3) × [(i√3) / (i√3)]
b = -19i√3 / (-6)
b = (19/6)i√3
[b(4 - i√3) / (4 + i√3)(4 - i√3)] - [b(4 + i√3) / (4 + i√3)(4 - i√3)] = 1
[(4b - bi√3) / (16 + 3)] - [(4b + bi√3) / (16 + 3)] = 1
[(4b - bi√3) - (4b + bi√3)] / 19 = 1
(4b - bi√3 - 4b - bi√3) / 19 = 1
-2bi(√3) / 19 = 1
b = (-19 / 2i√3) × [(i√3) / (i√3)]
b = -19i√3 / (-6)
b = (19/6)i√3
2018-02-06 10:07 am
[b/(4+i√3)] - [b/(4-i√3)] = 1
.......b............b
----------- - --------- = 1
(4+i√3)......(4-i√3)
// Use common denominator (4+i√3)(4-i√3)
b(4-i√3) - b(4+i√3)
------------------------- = 1
......(4+i√3)(4-i√3)
// Cross multiply and simplify
b(4-i√3) - b(4+i√3) = 1(4+i√3)(4-i√3)
4b - ib√3 - 4b - ib√3 = 16-3i²
- 2ib√3 = 16+3
- 2ib√3 = 19
.......19
b = ------...........................ANS
......-2i√3
or
.......19.......√3......-19√3
b = ------ • ------ = --------.............ANS
......-2i√3.....√3.......6i
.......b............b
----------- - --------- = 1
(4+i√3)......(4-i√3)
// Use common denominator (4+i√3)(4-i√3)
b(4-i√3) - b(4+i√3)
------------------------- = 1
......(4+i√3)(4-i√3)
// Cross multiply and simplify
b(4-i√3) - b(4+i√3) = 1(4+i√3)(4-i√3)
4b - ib√3 - 4b - ib√3 = 16-3i²
- 2ib√3 = 16+3
- 2ib√3 = 19
.......19
b = ------...........................ANS
......-2i√3
or
.......19.......√3......-19√3
b = ------ • ------ = --------.............ANS
......-2i√3.....√3.......6i
2018-02-06 10:06 am
b(4-i√3) - b(4+i√3)=19
- (2i√3)b=19
b = ...
- (2i√3)b=19
b = ...
2018-02-06 10:04 am
b * (1/(4 + i * sqrt(3)) - 1 / (4 - i * sqrt(3))) = 1
b * ((4 - i * sqrt(3) - 4 - i * sqrt(3)) / (4^2 - i^2 * 3)) = 1
b * (-2 * i * sqrt(3) / (16 + 3)) = 1
b * (-2 * i * sqrt(3) / 19) = 1
b = 19 / (-2 * i * sqrt(3))
b = -19 * sqrt(3) / (2 * 3 * i)
b = -19 * sqrt(3) / (6i)
b = -19 * sqrt(3) * i / (6i^2)
b = -19 * sqrt(3) * i / (-6)
b = 19 * sqrt(3) * i / 6
b * ((4 - i * sqrt(3) - 4 - i * sqrt(3)) / (4^2 - i^2 * 3)) = 1
b * (-2 * i * sqrt(3) / (16 + 3)) = 1
b * (-2 * i * sqrt(3) / 19) = 1
b = 19 / (-2 * i * sqrt(3))
b = -19 * sqrt(3) / (2 * 3 * i)
b = -19 * sqrt(3) / (6i)
b = -19 * sqrt(3) * i / (6i^2)
b = -19 * sqrt(3) * i / (-6)
b = 19 * sqrt(3) * i / 6
收錄日期: 2021-04-18 18:07:01
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