Представете в алгебричен вид числото

[zaiche10]

Здравейте. Може ли да ми помогнете със следните задачи. Ако може и обяснение. Благодаря ви искрено !





Става въпрос за последната задача.

[ammornil]

[tex]Z=(1+i)^{20}=? \\
A=1+i.1 \\
\hspace{24} R_A=1 \hspace{24} I_A=1\\
\hspace{24} |A|=\sqrt{(R_A)^2+(I_A)^2}=\sqrt{2} \\
\hspace{24} \phi_A=arctg(\frac{I_A}{R_A})=arctg(1)=\frac{\pi}{4} \\
\hspace{24} A=|A|.e^{i.\phi_A}=\sqrt{2}.e^{i\frac{\pi}{4}}\\
\vspace{10} \\
Z=A^{20}=\sqrt{2}^{20}.e^{i.20.\frac{\pi}{4}}=2^{\frac{20}{2}}.e^{i.5\pi}=2^{10}.e^{i.(\pi+4\pi)} \\
(1+i)^{20}=2^{10}.cos(\pi+4\pi)+i.2^{10}sin(\pi+4\pi)=2^{10}.cos\pi+i.2^{10}sin\pi\\
(1+i)^{20}=1024.(-1)+i.1024.0 \hspace{24} \Rightarrow (1+i)^{20}=-1024[/tex]
***************************************************************************************************************************
[tex]Z=(\frac{1}{2}-i\frac{\sqrt{3}}{2})^{21}=? \\
A=\frac{1}{2}-i\frac{\sqrt{3}}{2} \\
\hspace{24} R_A=\frac{1}{2} \hspace{24} I_A=-\frac{\sqrt{3}}{2}\\
\hspace{24} |A|=\sqrt{(R_A)^2+(I_A)^2}=1 \\
\hspace{24} \phi_A=arctg(\frac{I_A}{R_A})=arctg(-\sqrt{3})=-\frac{\pi}{3} \\
\hspace{24} A=|A|.e^{i.\phi_A}=1.e^{-i\frac{\pi}{3}}\\
\vspace{10} \\
Z=A^{21}=1^{21}.e^{-i.21.\frac{\pi}{3}}=1.e^{-i.7\pi}=1.e^{-i.(\pi+6\pi)} \\
(\frac{1}{2}-i\frac{\sqrt{3}}{2})^{21}=1.cos(\pi+6\pi)-i.1.sin(\pi+6\pi)=1.cos\pi-i.1sin\pi\\
(\frac{1}{2}-i\frac{\sqrt{3}}{2})^{21}=1.(-1)+i.1.0 \hspace{24} \Rightarrow (\frac{1}{2}-i\frac{\sqrt{3}}{2})^{21}=-1[/tex]
***************************************************************************************************************************
[tex]Z=(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})^{101}=? \\
A=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \\
\hspace{24} R_A=-\frac{\sqrt{2}}{2} \hspace{24} I_A=\frac{\sqrt{2}}{2}\\
\hspace{24} |A|=\sqrt{(R_A)^2+(I_A)^2}=1 \\
\hspace{24} \phi_A=arctg(\frac{I_A}{R_A})=arctg(-1)=-\frac{\pi}{4} \\
\hspace{24} A=|A|.e^{i.\phi_A}=1.e^{-i\frac{\pi}{4}}\\
\vspace{10} \\
Z=A^{101}=1^{101}.e^{-i.101.\frac{\pi}{4}}=1.e^{-i(\frac{\pi}{4}+100.\frac{\pi}{4})}=1.e^{-i.(\frac{\pi}{4}+\pi+24\pi)} \\
(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})^{101}=1.cos(\frac{\pi}{4}+\pi+24\pi)-i.1.sin(\frac{\pi}{4}+\pi+24\pi)=1.cos(\frac{\pi}{4}+\pi)-i.sin(\frac{\pi}{4}+\pi)=-1.cos(\frac{\pi}{4})+1.sin(\frac{\pi}{4}) \\
(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})^{101}=-1.(\frac{\sqrt{2}}{2})+i.1.\frac{\sqrt{2}}{2} \hspace{24} \Rightarrow (-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})^{101}=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}[/tex]

[ammornil]

[tex]\frac{1-i}{1+i}=\frac{1-i}{1+i}.\frac{1-i}{1-i}=\frac{(1-i)^2}{1^2-i^2}=\frac{1-2i+i^2}{1-(-1)}=\frac{1-2i-1}{2}=-i[/tex]

Знаем, че ако [tex]n \in N[/tex] то [tex]\left{i^{4n}=1 \\ i^{4n+1}=i \\ i^{4n+2}=-1 \\ i^{4n+3}=-i \right[/tex]

В случая:
[tex](\frac{1-i}{1+i})^{2002}=(-i)^{2002}=i^{2002}=i^{4.500+2}=-1[/tex]