komplexní čísla
Ahoj, spočítal by někdo tyto příklady a napsal sem i postup? Nemůžu na to přijít
2+4i
----- .(2-i)
1+i
a další
2-4i
----- +(1+2i)² . i³
1+i
2+4i
----- .(2-i)
1+i
a další
2-4i
----- +(1+2i)² . i³
1+i
Odpovědi
Diskuze
2+4i
----- .(2-i) =
1+i
(2+4i)*(2-i)/(1+i) = ( (4+8i) - i*(2+4i) )/(1+i) = (8+6i)/(1+i) = (8+6i)/(1+i) =
(8+6i)/(1+i) * (1-i)/(1-i) = (8+6i)*(1-i)/(1-i*i) =
(8+6i)*(1-i)/(1+1) = ((8+6i)-i(8+6i))/2 = (14-2i)/2 =
= 7-i
=====
https://www.wolframalpha.com/input/?i=(2%2B4i)*(2-i)%2F(1%2Bi)+%3D
2-4i
----- +(1+2i)² . i³ = A + B
1+i
A + B = (2-4i)/(1+i) + (1+2i)² * i³ =
A = (2-4i)/(1+i) = (2-4i)/(1+i) * 1 = (2-4i)/(1+i) * (1-i)/(1-i) =
(2-4i)*(1-i)/(1+1) = ( (2-4i) - i*(2-4i) )/2 = (-2-6i)/2 = (-1-3i) =
B = (1+2i)² * i³ = (1 + 2*2i + (2i)²) * (-i) = (1 + 4i - 4 ) * (-i) = (3i + 4)
A + B = (-1-3i) - (3i + 4) =
= 3
===
https://www.wolframalpha.com/input/?i=(2-4i)%2F(1%2Bi)+%2B+(1%2B2i)%C2%B2+*+i%C2%B3+%3D