Odpowiedź:
[tex]$z^{*}=\frac{\sqrt{3}-1}{16384 } +\frac{\sqrt{3}+1}{16384} i[/tex]
Szczegółowe wyjaśnienie:
Korzystamy ze wzoru de Moivre'a :
[tex]z^{15}=(1-i)^{15} \iff z = 1-i[/tex]
[tex]|z|=\sqrt{1^{2}+(-1)^{2}} =\sqrt{2}[/tex]
[tex]$\sin \varphi=\frac{b}{|z|} =-\frac{\sqrt{2}}{2}[/tex]
[tex]$\cos \varphi=\frac{a}{|z|} =\frac{\sqrt{2}}{2}[/tex]
Stąd:
[tex]$\varphi=\frac{7\pi}{4}[/tex]
Zatem:
[tex]$z=\sqrt{2}\Big(\cos\Big(\frac{7\pi}{4} \Big)+i \sin \Big(\frac{7\pi}{4} \Big) \Big)[/tex]
[tex]$z^{15}=(\sqrt{2})^{15}\Big(\cos \Big(\frac{105\pi}{4} \Big)+i \sin \Big(\frac{105 \pi}{4} \Big) \Big)=128\sqrt{2}\Big(\cos \Big(\frac{\pi}{4} \Big)+i \sin \Big(\frac{\pi}{4} \Big) \Big)=[/tex]
[tex]$=128\sqrt{2} \Big(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i \Big)=128(1+i)[/tex]
Teraz:
[tex]w^{20}=(1+i\sqrt{3})^{20} \iff w = 1+i\sqrt{3}[/tex]
[tex]$|w|=\sqrt{1^{2}+(\sqrt{3})^{2}} =\sqrt{4}=2[/tex]
[tex]$\sin \varphi=\frac{b}{|w|}=\frac{\sqrt{3}}{2}[/tex]
[tex]$\cos \varphi=\frac{a}{|w|}=\frac{1}{2}[/tex]
Stąd:
[tex]$\varphi =\frac{\pi}{3}[/tex]
Zatem:
[tex]$w=2\Big(\cos \Big(\frac{\pi}{3} \Big)+i \sin \Big(\frac{\pi}{3} \Big) \Big)[/tex]
[tex]$w^{20}=2^{20}\Big(\cos \Big(\frac{20\pi}{3} \Big)+i \sin \Big(\frac{20\pi}{3} \Big) \Big)=2^{20}\Big(\cos \Big(\frac{2\pi}{3} \Big)+i \sin \Big(\frac{2\pi}{3} \Big)\Big)=[/tex]
[tex]$=2^{20}\Big(-\frac{1}{2} +\frac{\sqrt{3}}{2}i \Big)[/tex]
Zatem:
[tex]$\frac{(1-i)^{15}}{(1+i\sqrt{3})^{20}} =\frac{2^{7}(1+i)}{2^{20}\Big(-\frac{1}{2}+\frac{\sqrt{3}}{2}i \Big)} =\frac{1+i}{2^{13}\Big(-\frac{1}{2}+\frac{\sqrt{3}}{2}i \Big)} =\frac{1+i}{-4096+4096i\sqrt{3}} =[/tex]
[tex]$=-\frac{1}{4096} \cdot \frac{1+i}{1-i\sqrt{3}} =-\frac{1}{4096} \cdot \frac{(1+i)(1+i\sqrt{3})}{(1-i\sqrt{3})(1+i\sqrt{3})} =[/tex]
[tex]$=-\frac{1}{4096} \cdot \frac{1+i\sqrt{3}+i-\sqrt{3}}{4} =-\frac{1}{4096} \Big(\frac{1-\sqrt{3}}{4} +\frac{1+\sqrt{3}}{4}i \Big)=[/tex]
[tex]$=\frac{\sqrt{3}-1}{16384 } -\frac{\sqrt{3}+1}{16384} i[/tex]
Liczba sprzężona:
[tex]$z^{*}=\frac{\sqrt{3}-1}{16384 } +\frac{\sqrt{3}+1}{16384} i[/tex]
