Liczby zespolone
Dominik: Cześć. Jak to policzyć?
| (1−3i)3 | | 1+ 9i+ 27i2+ 27i3 | |
| = |
| = ? |
| 2+2i | | 2+ 2i | |
23 lis 16:21
Saizou:
Skorzystaj z
i2=−1
i3=−i
23 lis 16:33
Dominik: ok,
| −26−18i | | 2−2i | | −52+52i−36i+36i2 | | −88 | | 16 | |
| * |
| = |
| = |
| + |
| i = −11+2i Gdzie |
| 2+2i | | 2−2i | | 4−4i2 | | 8 | | 8 | |
jest błąd?
23 lis 16:55
ABC: (−11+2i)(2+2i)=−22+4i−22i+4i
2=−26−18i nie ma błedu
23 lis 17:03
Dominik: a w odp mam −2+11i..
23 lis 17:08
Pytający:
Błąd w pierwszym poście:
| | | | | | | | | |
(1−3i)3= | 13*(−3i)0+ | 12*(−3i)1+ | 11*(−3i)2+ | 10*(−3i)3= |
| | | | | |
=1−9i+27i
2−27i
3=−26+18i
23 lis 18:20
Mila:
| | (1−3i)2*[(1−3i)*(2−2i)] | |
= |
| = |
| | (2+2i)*(2−2i) | |
| | 1−6i+9i2)*(2−2i−6i+6i2) | |
= |
| = |
| | 4−4i2 | |
| | (−8−6i) *(−8i−4) | | 2*(−4−3i)*4*(−2i−1) | |
= |
| = |
| = |
| | 8 | | 8 | |
=(4+3i)*(1+2i)=4+8i+3i+6i
2=−2+11i
========================
23 lis 18:36
