Комплексни бројеви 6

Текст задатака објашњених у видео лекцији.

Пр.13   Степеновати дати комплексни број.

             $z = 2 + 3i$      ${z^2} = ?$   ${z^3} = ?$

Пр.14   Одреди степен од ${z^{50}}$ датог комплексног броја.

             $z = 1 - i$      ${z^{50}} = ?$

Пр.15   Израчунај вресност следећег израза:

            $\frac{{{{\left( {1 + i} \right)}^{13}}}}{{{{\left( {1 - i} \right)}^7}}} = $

Пр.13

\[\begin{gathered}
z = 2 + 3i \hfill \\
{z^2} = {\left( {2 + 3i} \right)^2} = {2^2} + 2 \cdot 2 \cdot 3i + {\left( {3i} \right)^2} = 4 + 12i + 9{i^2} = \hfill \\
= 4 + 12i - 9 = - 5 + 12i \hfill \\
{z^3} = {\left( {2 + 3i} \right)^3} = {2^3} + 3 \cdot {2^2} \cdot 3i + 3 \cdot 2 \cdot {\left( {3i} \right)^2} + {\left( {3i} \right)^3} = \hfill \\
= 8 + 36i + 6 \cdot 9{i^2} + 27{i^3} = 8 + 36i - 54 - 27i = - 46 + 9i \hfill \\
\end{gathered} \]

Пр.14

\[\begin{gathered}
z = 1 - i \hfill \\
{z^{50}} = {\left( {1 - i} \right)^{50}} = {\left( {{{\left( {1 - i} \right)}^2}} \right)^{25}} = {\left( {1 - 2i - 1} \right)^{25}} = {\left( { - 2i} \right)^{25}} = \hfill \\
= - {2^{25}} \cdot {i^{6 \cdot 4 + 1}} = - {2^{25}} \cdot i \hfill \\
\end{gathered} \]

Пр. 15

\[\begin{gathered}
\frac{{{{\left( {1 + i} \right)}^{13}}}}{{{{\left( {1 - i} \right)}^7}}} = \frac{{\left( {1 + i} \right){{\left( {1 + i} \right)}^{12}}}}{{\left( {1 - i} \right){{\left( {1 - i} \right)}^6}}} = \frac{{\left( {1 + i} \right){{\left( {{{\left( {1 + i} \right)}^2}} \right)}^6}}}{{\left( {1 - i} \right){{\left( {{{\left( {1 - i} \right)}^2}} \right)}^3}}} = \hfill \\
= \frac{{\left( {1 + i} \right){{\left( {1 + 2i + {i^2}} \right)}^6}}}{{\left( {1 - i} \right){{\left( {1 - 2i + {i^2}} \right)}^3}}} = \frac{{\left( {1 + i} \right){{\left( {2i} \right)}^6}}}{{\left( {1 - i} \right){{\left( { - 2i} \right)}^3}}} = \frac{{\left( {1 + i} \right){2^6}{i^6}}}{{\left( {1 - i} \right){{\left( { - 2} \right)}^3}{i^3}}} = \hfill \\
= \frac{{\left( {1 + i} \right)8\left( { - i} \right)}}{{ - \left( {1 - i} \right)}} = \frac{{8i\left( {1 + i} \right)}}{{\left( {1 - i} \right)}} = \frac{{8i\left( {1 + i} \right)}}{{\left( {1 - i} \right)}} \cdot \frac{{1 + i}}{{1 + i}} = \frac{{8i{{\left( {1 + i} \right)}^2}}}{{1 - {i^2}}} = \hfill \\
= \frac{{8i\left( {1 + 2i + {i^2}} \right)}}{2} = \frac{{16{i^2}}}{2} = - 8 \hfill \\
\end{gathered} \]