The problem is: (-3 + i) / -3i
Thanks for the help!
2007-11-30 18:38:47 · 4 answers · asked by sillyboys_trucksare4girls 2 in Science & Mathematics ➔ Mathematics
(-3 + i) (3 i) / (- 3 i) (3 i)
(- 9 i - 3) / 9
(- 3) (1 + 3 i) / 9
(- 1/3) (1 + 3 i)
2007-11-30 21:54:27 · answer #1 · answered by Como 7 · 3⤊ 1⤋
In general, since i^2 = -1,
(a + bi)(a - bi) = a^2 + b^2
so
1/(a + bi) = (a - bi)/(a^2 + b^2).
Using this, you can turn any division into a multiplication, for example:
(2 + 3i)/(1 + i) = (2 + 3i)(1 - i)/2
or for your problem,
(-3 + i)/(-3i) = (-3 + i)(3i)/9
2007-12-01 02:55:09 · answer #2 · answered by a²+b²=c² 4 · 1⤊ 2⤋
mltiply the numerator by -3i
2007-12-01 03:26:00
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answer #3
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answered by gm_2422 2
·
1⤊
1⤋
so the result is::
9i^-3i^2 <
(-3 + i) / -3i
= [(-3 + i) / -3i ](i/i)
= (-3i - 1)/3
= -1/3 - i
2007-12-01 02:41:23 · answer #4 · answered by sahsjing 7 · 3⤊ 0⤋