Odpowiedź:
[tex]-1[/tex]
Szczegółowe wyjaśnienie:
Niech:
[tex]$\frac{z^{5}}{w^{3}}= \frac{\Big(\cos \frac{9}{17}\pi +i \sin \frac{9}{17}\pi \Big)^{5} }{\Big(\cos \frac{2}{17}\pi -i \sin \frac{2}{17}\pi \Big)^{3}}[/tex]
Będziemy korzystali ze wzoru de Moivre'a. Mamy:
[tex]$z^{5}=\Big(\cos \frac{9}{17}\pi +i \sin \frac{9}{17}\pi \Big)^{5} =\cos \frac{45}{17}\pi +i \sin \frac{45}{17}\pi=\cos \frac{11}{17}\pi +i \sin \frac{11}{17}\pi=[/tex]
[tex]$=e^{\frac{11}{17}\pi i }[/tex]
[tex]$w^{3}=\Big(\cos \frac{2}{17}\pi -i \sin \frac{2}{17}\pi\Big)^{3}=\cos \frac{6}{17} \pi -i \sin \frac{6}{17} \pi =e^{-\frac{6}{17}\pi i }[/tex]
Zatem:
[tex]$\frac{\Big(\cos \frac{9}{17}\pi +i \sin \frac{9}{17}\pi \Big)^{5} }{\Big(\cos \frac{2}{17}\pi -i \sin \frac{2}{17}\pi \Big)^{3}}=\frac{e^{\frac{11}{17}\pi i }}{e^{-\frac{6}{17}\pi i }} =e^{\pi i}=-1[/tex]
