a. 1/3 + 2/3i
b. 1/5 + 2/5i
c. 1/3 + i
d. 1/5 + 3/5i
2007-03-21 03:12:50 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1 +i / 2 -i
= (1 + i) (2 + i)/ (2 - i)(2 + i)
= (2 + i^2 + 3i) / (4 - i^2)
= (2 -1 + 3i) / (4 - -1) where i^2 = -1
= (1 + 3i) / 5
= 1/5 + 3/5i
(answer d)
2007-03-21 03:18:24 · answer #1 · answered by seah 7 · 1⤊ 0⤋
To do this you need to multiple the numerator and denominator by the denominator complex number conjugate (i.e (2+i))
((1+i)(2+i))/((2-i)(2+i))
= (2 + i +2i +i^2)/(4- i^2)
But i^2 = -1
( 2 + 3i - 1)/(4 + 1)
(1 + 3i)/5
= 1/5 + (3/5) I
Answer d.
2007-03-21 10:21:05 · answer #2 · answered by The exclamation mark 6 · 1⤊ 0⤋
(1 + i) / (2 - i)
Multiply the top and bottom by the conjugate of the denominator
[(1 + i) / (2 - i)] * [(2 + i) / (2 + i)]
[2 + 3i + i^2] / [4 - i^2]
[2 + 3i - 1] / [4 + 1]
[1 + 3i] / 5
1/5 + (3/5)i
Answer: d
2007-03-21 10:17:19 · answer #3 · answered by Bhajun Singh 4 · 1⤊ 0⤋
1+i / 2-i
Multiply top and bottom by (2+i):
(1+i)(2+i) / (2-i)(2+i)
= 2 + 3i - 1 / (4 + 1)
= 1/5 + 3i/5
D
2007-03-21 10:17:38 · answer #4 · answered by MamaMia © 7 · 0⤊ 0⤋
1+i / 2-i
= [(1+i)(2+i)] / [(2-i)(2+i)]
= (2+i+2i+i^2) / (4+2i-2i-i^2)
= (2+3i-1) / [4-(-1)]
= 1+3i / 5 (d)
2007-03-21 10:25:36 · answer #5 · answered by XxSyNd3rXx 2 · 0⤊ 0⤋