7 - 3i over 5 + 4i
2006-12-10 06:12:15 · 6 answers · asked by crazylifer 3 in Science & Mathematics ➔ Mathematics
(7-3i)(5-4i)/(5+4i)(5-4i)
=35-15i-28i-12/25+16
=(23/41)-(43/41)i
2006-12-10 06:15:44 · answer #1 · answered by raj 7 · 0⤊ 1⤋
you mutiply numerator and denominator by conjugate complex of denominator
the conjugate of 5+4i is 5-4i
so (7-3i) (5-4i) = 25 +16 =41 (comes from (a-b) (a+b) = a^2-b^2
and don't forget that i^2 = -1!!
developping numerator gives 35 -15i -28i +12 = 47-43i
so the result
47/41 - 43 i /41
2006-12-10 06:19:39 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
yea... first u multipty everything by 5-4i/5-4i.. then u get rid of i's.. do the rest
2006-12-10 06:14:06 · answer #3 · answered by coolchess123 3 · 0⤊ 0⤋
rationalize 7i(-9 +14i) / 7i(7i) (-63i+ 98i^2) / 49i^2 word that i^2 = -a million (-63i +ninety 8(-a million)) / 40 9(-a million) 7 (-9i -14) / -40 9 word that 7 is going into 7 as quickly as and into 40 9 seven circumstances -(9i +14) / -7 9i/7 + 14/7 9i/7 + 2 2 + 9i/7
2016-10-14 10:00:55 · answer #4 · answered by ? 4 · 0⤊ 0⤋
4 +1bi
2006-12-10 06:14:17 · answer #5 · answered by Anonymous · 0⤊ 0⤋
4
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9
2006-12-10 06:15:43 · answer #6 · answered by ashantiea 1 · 0⤊ 0⤋