Liczby zespolone
Zl: Przedstaw w postaci trygonometrycznej:
a) (1+i)(cosα+ isinα)
b) 1−cosα−isinα
25 lis 14:06
grzest:
a)
(1+i)(cosα+ isinα)=
=(1+i)eiα=21/2eiπ/4eiα=√2ei(α+π/4)=√2(cos(α+π/4)+isin(α+π/4)).
z=1+i
r=√2, cos φ=1/√2, sin φ =1/√2 ⇒ φ=π/4.
25 lis 14:43
grzest:
b)1−cosα−isinα= 1−eiα =
(1−eiα/2)(1+eiα/2)=(1−cos(α/2)−isin(α/2))(1+cos(α/2)+isin(α/))=
=2sin(α/2)[sin(α/2)−icos(α/2)].
25 lis 15:16
Mila:
a)z
1*z
2=(1+i)(cosα+ isinα)=
| | π | | π | |
=√2*(cos |
| +i sin |
| )*1*(cosα+i sinα)= |
| | 4 | | 4 | |
| | π | | π | |
=√2*(cos(α+ |
| )+i sin(α+ |
| )) |
| | 4 | | 4 | |
25 lis 21:25