1−i√3 | ||
( | )37 | |
2 |
√3 | 1 | |||
wyszlo mi | + | i | ||
2 | 2 |
5 | 5 | |||
po spotegowaniu we wzorze mam 1*(cos ( | π +isin( | π)) | ||
6) | 6 |
π | ||
potem do zrobilem do 4 cwiartki czyli π−a0 przyjalem π− | ||
6 |
1 | √3 | π | π | 5π | 5π | ||||||
− | i = cos(2π− | ) + i sin(2π− | ) = cos | + i sin | |||||||
2 | 2 | 3 | 3 | 3 | 3 |
1 | √3 | 185π | 185π | |||||
( | − | )37 = cos | + i sin | = | ||||
2 | 2 | 3 | 3 |
5π | 5π | 5π | 5π | |||||
= cos(60π+ | ) + i sin(60π+ | ) = cos( | ) + i sin( | ) = | ||||
3 | 3 | 3 | 3 |
1 | √3 | |||
= | − | i | ||
2 | 2 |