\(\frac{(1+i)^{21}}{(1-i)^{19}}\)
nie mogę napisać potęgi ale w liczniku jest do 21, w mianowniku do 19
obliczyć l. zespolone
Otrzymałeś(aś) rozwiązanie do zamieszczonego zadania? - podziękuj autorowi rozwiązania! Kliknij
\(z_1=1+i\\r=\sqrt{1^2+1^2}=\sqrt{2}\\z_1=\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)=\sqrt{2}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))\)
\(z_1^{21}=(\sqrt{2})^{21}(cos(21\cdot\frac{\pi}{4})+i\ sin(21\cdot\frac{\pi}{4}))\\=(\sqrt{2})^{21}(cos(\frac{21}{4}\pi)+i\ sin(\frac{21}{4}\pi))=\\=(\sqrt{2})^{21}(cos(\frac{5}{4}\pi)+i\ sin(\frac{5}{4}\pi))\)
\(z_2=1-i\\r_2=\sqrt{1^2+(-1)^2}=\sqrt{2}\\z_2=\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)=\sqrt{2}(cos(\frac{7}{4}\pi)+i\ sin(\frac{7}{4}\pi))\)
\(z_2^{19}=(\sqrt{2})^{19}(cos(19\cdot\frac{7}{4}\pi)+i\ sin(19\cdot\frac{7}{4}\pi))=\\=(\sqrt{2})^{19}(cos(\frac{133}{4}\pi)+i\ sin(\frac{133}{4}\pi))=\\=(\sqrt{2})^{19}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))\)
\(\frac{z_1^{21}}{z_2^{19}}=\frac{(\sqrt{2})^{21}}{(\sqrt{2})^{19}}(cos(\frac{5}{4}\pi-\frac{\pi}{4})+i\ sin(\frac{5}{4}\pi-\frac{\pi}{4}))=\\=(\sqrt{2})^2(cos\pi+i\ sin\pi)=2(-1+i\cdot0)=2(-1)=-2\)
\(z_1^{21}=(\sqrt{2})^{21}(cos(21\cdot\frac{\pi}{4})+i\ sin(21\cdot\frac{\pi}{4}))\\=(\sqrt{2})^{21}(cos(\frac{21}{4}\pi)+i\ sin(\frac{21}{4}\pi))=\\=(\sqrt{2})^{21}(cos(\frac{5}{4}\pi)+i\ sin(\frac{5}{4}\pi))\)
\(z_2=1-i\\r_2=\sqrt{1^2+(-1)^2}=\sqrt{2}\\z_2=\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)=\sqrt{2}(cos(\frac{7}{4}\pi)+i\ sin(\frac{7}{4}\pi))\)
\(z_2^{19}=(\sqrt{2})^{19}(cos(19\cdot\frac{7}{4}\pi)+i\ sin(19\cdot\frac{7}{4}\pi))=\\=(\sqrt{2})^{19}(cos(\frac{133}{4}\pi)+i\ sin(\frac{133}{4}\pi))=\\=(\sqrt{2})^{19}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))\)
\(\frac{z_1^{21}}{z_2^{19}}=\frac{(\sqrt{2})^{21}}{(\sqrt{2})^{19}}(cos(\frac{5}{4}\pi-\frac{\pi}{4})+i\ sin(\frac{5}{4}\pi-\frac{\pi}{4}))=\\=(\sqrt{2})^2(cos\pi+i\ sin\pi)=2(-1+i\cdot0)=2(-1)=-2\)
-
Galen
- Guru
- Posty: 18457
- Rejestracja: 17 sie 2008, 15:23
- Podziękowania: 4 razy
- Otrzymane podziękowania: 9161 razy
A może trochę prościej:
\(\frac{(1+i)^{21}}{(1-i)^{19}}= \frac{(1+i)^{19} \cdot (1+i)^2}{(1-i)^{19}}= (\frac{1+i}{1-i})^{19} \cdot (1+i)^2=
= [\frac{(1+i)(1+i)}{(1-i)(1+i)}]^{19} \cdot (1+i)^2= (\frac{1+2i+i^2}{1-i^2})^{19} \cdot (1+2i+i^2)=
=( \frac{1+2i-1}{1+1})^{19} \cdot (1+2i-1)=( \frac{2i}{2})^{19} \cdot 2i=i^{19} \cdot 2i=i \cdot 2i=2i^2=2 \cdot (-1)=-2\)
\(\frac{(1+i)^{21}}{(1-i)^{19}}= \frac{(1+i)^{19} \cdot (1+i)^2}{(1-i)^{19}}= (\frac{1+i}{1-i})^{19} \cdot (1+i)^2=
= [\frac{(1+i)(1+i)}{(1-i)(1+i)}]^{19} \cdot (1+i)^2= (\frac{1+2i+i^2}{1-i^2})^{19} \cdot (1+2i+i^2)=
=( \frac{1+2i-1}{1+1})^{19} \cdot (1+2i-1)=( \frac{2i}{2})^{19} \cdot 2i=i^{19} \cdot 2i=i \cdot 2i=2i^2=2 \cdot (-1)=-2\)
Wszystko jest trudne,nim stanie się proste.