How in the heck do I solve this problem!?!
-28-16i / 1-5i
Any help would really be appreciated!
2007-09-20 09:34:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I bet they want you to simplify it
(-28 -16i)/(1-5i)
multiply the top and bottom by
the denominator's conjugate.
in this case, it is (1+5i)
(-28-16i)(1+5i)/(1+25)
(-28 -140i -16i +80)/26
(52 - 156i)/26
2 - 6i
2007-09-20 09:43:36 · answer #1 · answered by Greg G 5 · 0⤊ 0⤋
Multiply through top and bottom by the conjugate of the denominator.
(-28 - 16i)(1 + 5i) = -28 -140i -16i + 80 = 52 - 156i
(1 - 5i)(1 + 5i) = 1 - 5i + 5i + 5 = 6
Now divide the top by the bottom.
52/6 - 26i
2007-09-20 09:42:09 · answer #2 · answered by PMP 5 · 0⤊ 0⤋
i = sqrt(-1) and i^2 = -1. If you rationalise the denominator, i term will be eliminated in the denominator.
(-28 - 16i)/(1- 5i)
multiply with (1 + 5i)/(1+5i)
(-28 - 16i)(1 + 5i)/(1 - 5i)(1 + 5i)
-28 - 140 i - 16i - 80i^2/(1 - 25i^2) =
-28 - 156i - 80(-1)/(1 - 25(-1))
-28 - 156i +80/(1 +25) =
(52 - 156i)/26
= 2 - 6i
2007-09-20 09:51:35 · answer #3 · answered by mohanrao d 7 · 0⤊ 0⤋
Are you just trying to reduce it? Try that whole thing multiplying by (1+5i)/(1+5i). The (1+5i) is the conjugate, as others have noted. The key is that since i^2 = -1, you get rid of the i's that are squared because of that, and due to the conjugate, you'll have -5i and 5i that cancel out by adding to 0.
2007-09-20 10:02:33 · answer #4 · answered by Unweird Al 1 · 0⤊ 0⤋
Multiply the numerator and the denominator by 1 + 5i.
2007-09-20 09:43:50 · answer #5 · answered by Tony 7 · 0⤊ 0⤋