Thanks
2007-12-18 13:36:33 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Greetings,
(i + 1)(3 + i)/(2 - 4i) = z
z = (3i - 1 + 3 + i)/(2 - 4i)
= (2 + 4i)/(2 - 4i)
= [(2 + 4i)^2/[(2 - 4i)(2 + 4i)]
=( 4 - 16 + 16i)/20
= (-12 +16i)/20
= -3/5 + 4i/5
Regards
2007-12-18 13:43:58 · answer #1 · answered by ubiquitous_phi 7 · 0⤊ 0⤋
Solve for z.
(1 + i) / (2 - 4i) = z / (3 + i)
(3 + i)(1 + i) / (2 - 4i) = z
(2 + 4i) / (2 - 4i) = z
(2 + 4i)² / [(2 - 4i)(2 + 4i)] = z
(-12 + 16i) / 20 = z
z = -12/20 + (16/20)i
z = -3/5 + (4/5)i
2007-12-18 21:54:13 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
z = (1+i)(3+i)/(2-4i) = (2+4i)/(2-4i) = (-12 + 16i)/20 =
(-3/5) + (4/5)i
2007-12-18 21:52:58 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(3+i)(1+i)=z(2-4i)
3+3i+i+i^2=z(2-4i)
z=3+4i+i^2/(2-4i)
2007-12-18 21:43:46 · answer #4 · answered by Anonymous · 0⤊ 0⤋