حل السؤالين 2و3 من التمرين الثاني
\begin{array}{l}
m = \sqrt 2 e^{i\theta } \\
\overline m = \sqrt 2 e^{ - i\theta } \\
\Delta = \left( { - 2} \right)^2 - 4\left( {m\overline m } \right) \\
= 4 - 4\left( {\sqrt 2 e^{i\theta } \sqrt 2 e^{ - i\theta } } \right) \\
= 4 - 4\left( {2e^{i\theta } ^{ - i\theta } } \right) \\
= 4 - 8 = - 4 \\
\sqrt \Delta = 2i \\
z_1 = \frac{{2 + 2i}}{{2m}} = \frac{{2 + 2i}}{{2\sqrt 2 e^{i\theta } }} = \frac{{2\sqrt 2 \left( {\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i} \right)}}{{2\sqrt 2 e^{i\theta } }} = \frac{{\left( {\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}i} \right)}}{{e^{i\theta } }} = \frac{{\left( {e^{i\left( {\frac{\pi }{4}} \right)} } \right)}}{{e^{i\theta } }} = e^{i\left( {\frac{\pi }{4} - \theta } \right)} \\
z_2 = \frac{{2 - 2i}}{{2m}} = \frac{{2 - 2i}}{{2\sqrt 2 e^{i\theta } }} = \frac{{2\sqrt 2 \left( {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \right)}}{{2\sqrt 2 e^{i\theta } }} = \frac{{\left( {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \right)}}{{e^{i\theta } }} = \frac{{\left( {e^{i\left( { - \frac{\pi }{4}} \right)} } \right)}}{{e^{i\theta } }} = e^{ - i\left( {\frac{\pi }{4} + \theta } \right)} \\
\end{array}
الصقي الحل في برنامج mathtype
اما الباقي الاسئلة اظن انها سهلة