Prijemni ispit ETF – 26. jun 2017.
4. zadatak
Dati su kompleksni brojevi [inlmath]z_1=2017+2018i[/inlmath], [inlmath]z_2=2018+2019i[/inlmath] i [inlmath]\displaystyle w=(z_2-z_1)^{-2020}\cdot\left(\frac{\overline{z_1}+1}{2018}\right)^{2021}[/inlmath] (gde je [inlmath]\overline{z_1}[/inlmath] konjugovano kompleksni broj broja [inlmath]z_1[/inlmath] i [inlmath]i^2=-1[/inlmath]). Tada je [inlmath]|w|[/inlmath] jednak:
[inlmath]\text{(A)}\;\sqrt5\hspace{1cm}\text{(B)}\;\sqrt{2017}\hspace{1cm}\text{(C)}\;\sqrt{2020}\hspace{1cm}\enclose{circle}{\text{(D)}\;\sqrt2}\hspace{1cm}\text{(E)}\sqrt{10}\hspace{1cm}\text{(N)}\;[/inlmath]Ne znam
[inlmath]z_1=2017+2018i\\
z_2=2018+2019i\\
\displaystyle w=(z_2-z_1)^{-2020}\cdot\left(\frac{\overline{z_1}+1}{2018}\right)^{2021}[/inlmath]
[dispmath](z_2-z_1)^{-2020}=(2018+2019i-(2017+2018i))=\\
=(2018+2019i-2017-2018i)=\\
=\enclose{box}{(1+i)^{-2020}}\\
\left(\frac{\overline{z_1}+1}{2018}\right)^{2021}=\left(\frac{2018-2018i}{2018}\right)^{2021}=\left(\frac{\cancel{2018}(1-i)}{\cancel{2018}}\right)^{2021}=\\
=\enclose{box}{(1-i)^{2021}}\\
(1+i)^{-2020}\cdot(1-i)^{2021}=\frac{1}{(1+i)^{2020}}\cdot(1-i)^{2021}=\frac{(1-i)^{2021}}{(1+i)^{2020}}=\\
=\enclose{box}{1-i}\\
|w|=\sqrt{1^2+1^2}=\enclose{box}{\sqrt2}[/dispmath]