Rozwiązanie:
Głównie korzystamy ze wzoru de Moivre'a.
[tex]a)[/tex]
[tex]$\frac{(1-i)^{24}}{(\sqrt{3}-i)^{22} }[/tex]
Mamy:
Dla [tex]1-i[/tex] :
[tex]|z|=\sqrt{1+1} =\sqrt{2}[/tex]
[tex]$sin\varphi=-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2} \wedge cos \varphi=\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2} \Rightarrow \varphi=-\frac{\pi}{4}[/tex]
Dla [tex]\sqrt{3} -i[/tex] :
[tex]|z|=\sqrt{3+1} =2[/tex]
[tex]$sin \varphi=-\frac{1}{2} \wedge cos \varphi=\frac{\sqrt{3} }{2} \Rightarrow \varphi=-\frac{\pi }{6}[/tex]
Zatem:
[tex]$(1-i)^{24}=2^{12}\Big(cos\Big(-\frac{24\pi}{4} \Big)+isin\Big(\frac{-24\pi }{4} \Big)\Big)=2^{12}(cos(-6\pi )+isin(-6\pi ))=2^{12}[/tex]
[tex]$(\sqrt{3}-i )^{22}=2^{22}\Big(cos\Big(\frac{-22\pi }{6} \Big)+isin\Big(\frac{-22\pi }{6} \Big)\Big)=2^{22}\Big(cos\Big(\frac{\pi }{3} \Big)+isin\Big(\frac{\pi }{3} \Big)\Big)=[/tex]
[tex]$=2^{22}\Big(\frac{1}{2}+\frac{\sqrt{3} }{2} i \Big)[/tex]
Stąd:
[tex]$\frac{(1-i)^{24}}{(\sqrt{3}-i)^{22} }=\frac{2^{12}}{2^{22}\Big(\frac{1}{2}+\frac{\sqrt{3} }{2} i \Big)}=\frac{1}{2^{10}\Big(\frac{1}{2}+\frac{\sqrt{3} }{2} i \Big)}} =\frac{1}{512(1+ i\sqrt{3} )}[/tex]
[tex]b)[/tex]
[tex]$\frac{(-1+i\sqrt{3} )^{36}}{(1+i)^{32}}[/tex]
Mamy:
Dla [tex]-1+i\sqrt{3}[/tex] :
[tex]|z|=\sqrt{1+3} =2[/tex]
[tex]$sin \varphi=\frac{\sqrt{3} }{2} \wedge cos \varphi=-\frac{1}{2} \Rightarrow \varphi=\frac{2\pi }{3}[/tex]
Dla [tex]1+i[/tex] :
[tex]|z|=\sqrt{1+1} =\sqrt{2}[/tex]
[tex]$sin \varphi=\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2} \wedge cos \varphi=\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2} \Rightarrow \varphi=\frac{\pi }{4}[/tex]
Zatem:
[tex]$(-1+i\sqrt{3} )^{36}=2^{36}\Big(cos\Big(\frac{72\pi }{3} \Big)+isin\Big(\frac{72\pi }{3}\Big)\Big)=2^{36}(cos(24\pi )+isin(24\pi ))=2^{36}[/tex]
[tex]$(1+i)^{32}=2^{16}\Big(cos\Big(\frac{32\pi }{4} \Big)+isin\Big(\frac{32\pi }{4} \Big)\Big)=2^{16}(cos(8\pi )+isin(8\pi ))=2^{16}[/tex]
Stąd:
[tex]$\frac{(-1+i\sqrt{3} )^{36}}{(1+i)^{32}}=\frac{2^{36}}{2^{16}} =2^{20}[/tex]
[tex]c)[/tex]
[tex]$\frac{(-1-i)^{20}}{(1-i\sqrt{3})^{12} }[/tex]
Mamy:
Dla [tex]-1-i[/tex] :
[tex]|z|=\sqrt{1+1} =\sqrt{2}[/tex]
[tex]$sin \varphi=-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2} \wedge cos \varphi=-\frac{1}{\sqrt{2} }=-\frac{\sqrt{2} }{2} \Rightarrow \varphi=-\frac{3\pi }{4}[/tex]
Dla [tex]$1-i\sqrt{3}[/tex] :
[tex]|z|=\sqrt{1+3} =2[/tex]
[tex]$sin \varphi=-\frac{\sqrt{3} }{2} \wedge cos \varphi=\frac{1}{2} \Rightarrow \varphi=-\frac{\pi }{3}[/tex]
Zatem:
[tex]$(-1-i)^{20}=2^{10}\Big(cos\Big(\frac{-60\pi }{4} \Big)+isin\Big(\frac{-60\pi }{4}\Big)\Big)=2^{10}(cos(-15\pi) +isin(-15\pi ))=[/tex]
[tex]$=-2^{10}[/tex]
[tex]$(1-i\sqrt{3} )^{12}=2^{12}\Big(cos\Big(-\frac{12\pi }{3} \Big)+isin\Big(-\frac{12\pi }{3}\Big)\Big)=2^{12}(cos(-4\pi )+isin(-4\pi ))=[/tex]
[tex]$=2^{12}[/tex]
Stąd:
[tex]$\frac{(-1-i)^{20}}{(1-i\sqrt{3})^{12} }=\frac{-2^{10}}{2^{12}} =-\frac{1}{4}[/tex]
[tex]d)[/tex]
[tex]$\Big(\frac{1+i\sqrt{3} }{1-i} \Big)^{20}[/tex]
Mamy:
Dla [tex]1+i\sqrt{3}[/tex] :
[tex]|z|=\sqrt{1+3} =2[/tex]
[tex]$sin \varphi=\frac{\sqrt{3} }{2} \wedge cos \varphi=\frac{1}{2} \Rightarrow \varphi=\frac{\pi }{3}[/tex]
Dla [tex]1-i[/tex] :
[tex]|z|=\sqrt{1+1} =\sqrt{2}[/tex]
[tex]$sin\varphi=-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2} \wedge cos \varphi=\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2} \Rightarrow \varphi=-\frac{\pi}{4}[/tex]
Zatem:
[tex]$(1+i\sqrt{3} )^{20}=2^{20}\Big(cos\Big(\frac{20\pi }{3} \Big)+isin\Big(\frac{20\pi }{3} \Big)\Big)=2^{20}\Big(cos\Big(\frac{2\pi }{3} \Big)+isin\Big(\frac{2\pi }{3} \Big)\Big)=[/tex]
[tex]$=2^{20}\Big(-\frac{1}{2} +\frac{\sqrt{3} }{2}i \Big)[/tex]
[tex]$(1-i)^{20}=2^{10}\Big(cos\Big(\frac{-20\pi }{4} \Big)+isin\Big(\frac{-20\pi }{4} \Big)\Big)=2^{10}(cos(-5\pi )+isin(-5\pi ))=[/tex]
[tex]$=-2^{10}[/tex]
Stąd:
[tex]$\Big(\frac{1+i\sqrt{3} }{1-i} \Big)^{20}=-\frac{2^{20}\Big(-\frac{1}{2} +\frac{\sqrt{3} }{2}i \Big)}{2^{10}}=2^{10} \cdot \Big(\frac{1}{2}-\frac{\sqrt{3} }{2} i\Big)=512(1-i\sqrt{3} )[/tex]
[tex]e)[/tex]
[tex]\Big(1+i\sqrt{3} \Big)^{2003}[/tex]
Mamy:
[tex]|z|=\sqrt{1+3} =2[/tex]
[tex]$sin \varphi=\frac{\sqrt{3} }{2} \wedge cos \varphi=\frac{1}{2} \Rightarrow \varphi=\frac{\pi }{3}[/tex]
Zatem:
[tex]$\Big(1+i\sqrt{3} \Big)^{2003}=2^{2003}\Big(cos\Big(\frac{2003\pi }{3} \Big)+isin\Big(\frac{2003\pi }{3} \Big)\Big)=[/tex]
[tex]$=2^{2003}\Big(cos\Big(\frac{5\pi }{3} \Big)+isin\Big(\frac{5\pi }{3} \Big)\Big)=2^{2003} \Big(\frac{1}{2}-\frac{\sqrt{3} }{2}i \Big)[/tex]
