Odpowiedź:
[tex]$\Big(\frac{1+i}{i-\sqrt{3}}\Big)^{15}=-\frac{1}{256} -\frac{i}{256}[/tex]
Szczegółowe wyjaśnienie:
[tex]$\Big(\frac{1+i}{i-\sqrt{3}}\Big)^{15}=\frac{(1+i)^{15}}{(i-\sqrt{3})^{15}}[/tex]
Skorzystamy ze wzoru de Moivre'a :
[tex]z^{n}=|z|^{n}\Big(\cos (n\varphi )+i \sin(n \varphi)\Big)[/tex]
Licznik:
[tex]z^{15}=(1+i)^{15} \iff z =1+i[/tex]
[tex]|z|=\sqrt{1^{2}+1^{2}}=\sqrt{2}[/tex]
[tex]$\sin \varphi =\frac{b}{|z|} =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2}[/tex]
[tex]$\cos \varphi =\frac{a}{|z|} =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2}[/tex]
Stąd:
[tex]$\varphi =\frac{\pi}{4}[/tex]
Zatem:
[tex]$(1+i)^{15}=(\sqrt{2})^{15}\Big(\cos \Big(\frac{15\pi}{4} \Big)+ i\sin \Big(\frac{15\pi}{4}\Big)\Big)=[/tex]
[tex]$128\sqrt{2} \Big(\cos \Big(-\frac{\pi}{4} \Big)+i \sin \Big(-\frac{\pi}{4}\Big)\Big)=128\sqrt{2}\Big(\frac{\sqrt{2} }{2} -\frac{\sqrt{2} }{2} i\Big)=128(1-i)[/tex]
Mianownik:
[tex]z^{15}=(i-\sqrt{3})^{15} \iff z=i-\sqrt{3}[/tex]
[tex]|z|=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{4} =2[/tex]
[tex]$\sin \varphi=\frac{b}{|z|}=\frac{1}{2}[/tex]
[tex]$\cos \varphi = \frac{a}{|z|}=-\frac{\sqrt{3}}{2}[/tex]
Stąd:
[tex]$\varphi =\frac{5\pi}{6}[/tex]
Zatem:
[tex]$(i-\sqrt{3})^{15}=2^{15}\Big(\cos \Big(\frac{75\pi}{6} \Big)+i \sin \Big(\frac{75\pi}{6} \Big)\Big)=[/tex]
[tex]$=2^{15}\Big(\cos \Big(\frac{\pi}{2} \Big)+ i \sin \Big(\frac{\pi}{2} \Big)\Big)=2^{15}\Big(0+i)\Big)=2^{15}i[/tex]
Ostatecznie:
[tex]$\frac{(1+i)^{15}}{(i-\sqrt{3})^{15}}=\frac{2^{7}(1-i)}{2^{15}i} =\frac{1-i}{2^{8}i} =-\frac{i+1}{256}[/tex]
