I understand a little about the i but i dont see how 2 solve
2007-04-17 06:45:00 · 4 answers · asked by ♥♪ Zee ☆♫ 4 in Science & Mathematics ➔ Mathematics
(-17-i)(-3-i)/(-3+i)(-3-i)
51+17i+3i-1/9+1 (bcoz i^2=1)
50+20i/10
5+2i
2007-04-17 06:50:33 · answer #1 · answered by miinii 3 · 0⤊ 0⤋
Multiply numerator and denominator by
-3-i(which is the conjugate
=(-17-i)(-3-i)/10 =(51+17i+3i-1)/10=5+2i
2007-04-17 13:51:09 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
Multiply above and below the line by what's below the line then make it equal to zero to solve for i:
(-17-i)(-3+i)=0
If the solution equals zero then -17-i=0 or -3+i=0
So i can be -17 or 3
2007-04-17 13:53:57 · answer #3 · answered by EaterOfTartanColouredSmarties 4 · 0⤊ 0⤋
= - (17 + i).(-3 - i) / (-3 - i).(- 3 - i)
= - (- 51 - 20i + 1) / (9 + 1)
= (50 + 20i) / 10
= 5 + 2i
2007-04-17 14:38:24 · answer #4 · answered by Como 7 · 0⤊ 0⤋