(7-3i)(3+i)
2007-12-09 18:55:19 · 7 answers · asked by *TiNK* 3 in Science & Mathematics ➔ Mathematics
21 + 7i - 9i - 3i²
21 - 2i + 3
24 - 2i
2007-12-09 19:52:33 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Multiplying imaginary or complex numbers
1) Multiply same as polynomials
(7 - 3i) ( 3 + i) = (7x3) + (7 x i) + (-3i x 3) + (-3i x i) =
21 + 7i - 9i - 3i^2 =
21 - 2i - 3i^2 =
Remember 1^2 = -1
21 - 2i - 3(-1) =
21 - 2i + 3 =
21 - 2i ===> the answer
ALWAYS write the final answer in the standard form ==> a + bi
2007-12-10 03:05:23 · answer #2 · answered by detektibgapo 5 · 0⤊ 0⤋
Imaginary numbers.
( 7 - 3i ) ( 3 + i )
= 21 - 9i + 7i - 3i^2
Remember that i^2 = -1
= 21 -2i + 3
=24 - 2i
2007-12-10 03:00:21 · answer #3 · answered by Hk 4 · 0⤊ 0⤋
(7-3i)(3+i) = 21 +7i -9i -3i²
= 21-3(-1) + (7-9)i
= (21+3) + (7-9)i
= 24 -2i
2007-12-10 03:00:04 · answer #4 · answered by smci 7 · 0⤊ 1⤋
distribute like you would usually
7*3 + 7i - 9i - 3i^2
= 21 - 2i - 3i^2
you know that i^2 = -1
therefore:
= 21 - 2i +3
= 24 - 2i
2007-12-10 03:00:38 · answer #5 · answered by Zmik 3 · 0⤊ 0⤋
(7-3i)(3+i)
=7*3-3i*3+7*i-3i*i
=21-9i+7i+3
=24-2i
2007-12-10 03:00:20 · answer #6 · answered by Ian 6 · 0⤊ 0⤋
(7-3i)(3+i)
Use foil and get this equation:
21 - 9i + 7i - (3(i^2))
i^2 equals (-1)
*Final answer : 21-2i+3 = 24-2i
2007-12-10 03:03:32 · answer #7 · answered by Anonymous · 0⤊ 0⤋