razor: x
2−5x+7+i=0
Δ = 25−4(7+i) = 25−28−4i = −3−4i
√Δ =
√−3−4i
√−3−4i = a+bi
−3−4i = a
2+2abi−b
2
a
2−b
2 = −3
2ab = −4
a
2+b
2 = 5
2a
2 = 2
a = 1 lub a = −1
b = −2 lub b = 2
√−3−4i = {1−2i, −1+2i}
| | 5+1−2i | | 6−2i | |
x1 = |
| = |
| = 3−i |
| | 2 | | 2 | |
| | 5−1+2i | | 4+2i | |
x2 = |
| = |
| = 2+i |
| | 2 | | 2 | |
Re(3−i) = 3, Im(3−i) = −1
Re(2+i) = 2, Im(2+i) = 1