Комплексни бројеви 5

Текст задатака објашњених у видео лекцији.

Пр.9   Одреди количник датих комплексних бројева.

           ${z_1} =  - 1 + 3i$   ${z_2} = 2i$      ${z_1}:{z_2} = ?$

Пр.10  Одреди количник датих комплексних бројева.

            ${z_1} =  - 3 + 2i$   ${z_2} = 1 - i$      ${z_1}:{z_2} = ?$

Пр.11  Упростити следећи израз:

           $\frac{{1 - 2\sqrt 3 i}}{{2 + \sqrt 3 i}} = $

Пр.12  Израчунај вредност следећег израза:

            $\frac{1}{{1 + i}} - \frac{2}{{2 - \sqrt 2 i}} = $

Пр.9

\[\begin{gathered}
{z_1}{\text{ }} = {\text{ }} - {\text{ }}1{\text{ }} + {\text{ }}3i \hfill \\
{{\text{z}}_2}{\text{ }} = {\text{ }}2i \hfill \\
{z_1}:{z_2}{\text{ }} = \frac{{{z_1}}}{{{z_2}}} = \frac{{ - {\text{ }}1{\text{ }} + {\text{ }}3i}}{{2i}} \cdot \frac{i}{i} = \frac{{ - i + 3{i^2}}}{{2{i^2}}} = \frac{{ - i - 3}}{{ - 2}} = \hfill \\
= \frac{{ - i}}{{ - 2}} - \frac{3}{2} = - \frac{3}{2} + \frac{1}{2}i \hfill \\
\end{gathered} \]

Пр. 10

\[\begin{gathered}
{z_1}{\text{ }} = {\text{ }} - {\text{ }}3{\text{ }} + {\text{ }}2i \hfill \\
{{\text{z}}_2}{\text{ }} = {\text{ }}1{\text{ }} - {\text{ }}i \hfill \\
{z_1}:{z_2}{\text{ }} = \frac{{{z_1}}}{{{z_2}}} = \frac{{ - {\text{ }}3{\text{ }} + {\text{ }}2i}}{{1{\text{ }} - {\text{ }}i}} \cdot \frac{{1{\text{ }} + {\text{ }}i}}{{1{\text{ }} + {\text{ }}i}} = \frac{{ - 3 - 3i + 2i + 2{i^2}}}{{{1^2} - {i^2}}} = \frac{{ - 3 - i - 2}}{2} = \hfill \\
= \frac{{ - 5 - i}}{2} = - \frac{5}{2} - \frac{1}{2}i \hfill \\
\end{gathered} \]

Пр. 11

\[\begin{gathered}
\frac{{1 - 2\sqrt 3 i}}{{2 + \sqrt 3 i}} = \frac{{1 - 2\sqrt 3 i}}{{2 + \sqrt 3 i}} \cdot \frac{{2 - \sqrt 3 i}}{{2 - \sqrt 3 i}} = \frac{{2 - \sqrt 3 i - 4\sqrt 3 i + 2\sqrt 3 \sqrt 3 {i^2}}}{{{2^2} - {{\left( {\sqrt 3 i} \right)}^2}}} = \hfill \\
= \frac{{2 - 5\sqrt 3 i + 6{i^2}}}{{4 - 3{i^2}}} = \frac{{2 - 5\sqrt 3 i - 6}}{{4 + 3}} = \frac{{ - 4 - 5\sqrt 3 i}}{7} = - \frac{4}{7} - \frac{{5\sqrt 3 }}{7}i \hfill \\
\end{gathered} \]

Пр. 12

\[\begin{gathered}
\frac{1}{{1 + i}} - \frac{2}{{2 - \sqrt 2 i}} = \frac{1}{{1 + i}} \cdot \frac{{1 - i}}{{1 - i}} - \frac{2}{{2 - \sqrt 2 i}} \cdot \frac{{2 + \sqrt 2 i}}{{2 + \sqrt 2 i}} = \hfill \\
= \frac{{1 - i}}{{1 - {i^2}}} - \frac{{4 - 2\sqrt 2 i}}{{{2^2} - {{\left( {\sqrt 2 i} \right)}^2}}} = \frac{{1 - i}}{{1 + 1}} - \frac{{4 - 2\sqrt 2 i}}{{{2^2} - 2{i^2}}} = \hfill \\
= \frac{{1 - i}}{2} - \frac{{4 - 2\sqrt 2 i}}{{4 + 2}} = \frac{{1 - i}}{2} - \frac{{4 - 2\sqrt 2 i}}{6} = \frac{{3 - 3i - \left( {4 - 2\sqrt 2 i} \right)}}{6} = \hfill \\
= \frac{{3 - 3i - 4 + 2\sqrt 2 i}}{6} = \frac{{ - 1 - 3i + 2\sqrt 2 i}}{6} = - \frac{1}{6} + \frac{{ - 3 + 2\sqrt 2 }}{6}i \hfill \\
\end{gathered} \]