أسئلة متنوعة

الملف الشخصي · 10-31-2014, 11:07 PM

(1) جد الحلول في مجموعة الأعداد المركبة للمعادلة الآتية :

${\color{Blue} \left ( z+1 \right )^{7}+\left ( z+2 \right )^{7}=0}$

(2) حلل المقدار الآتي الى أبسط صورة ممكنة بحيث لايحتوي على الوحدة التخيلية ( i ) :

${\color{Blue} \left ( z^{5} -32\right )}$

الملف الشخصي · 11-05-2014, 11:56 PM

حل السؤال الثاني :

$Let \ \ {\color{Blue} \left (z^{5}-32 \right )=0......\left ( 1 \right )}$

$\\ \ \ {\color{Blue} z^{5}-32=0}\Rightarrow{\color{Blue} z^{5}=32=32\left ( \cos \left ( 0 \right )+i\sin \left ( 0 \right ) \right )}\Rightarrow{\color{Blue}z=\left (32 \left (\cos \left ( 0 \right ) +i\sin \left ( 0 \right ) \right ) \right )^{\frac{1}{5}}}\\\Rightarrow {\color{Blue} z=\left (32 \right )^{\frac{1}{5}}\left ( \cos \left ( \frac{0+2k\pi }{5} \right ) +i\sin \left ( \frac{0+2k\pi }{5} \right )\right )}\ \ \ \ ;{\color{Blue} k=0,1,2,3,4}\\ \Rightarrow {\color{Blue} z=2\left ( \cos \left ( \frac{2k\pi }{5} \right )+i\sin \left ( \frac{2k\pi }{5} \right ) \right )}\ \ \ ;{\color{Blue} k=0,1,2,3,4}$

$\\ if\ \ {\color{Blue} k=0}\Rightarrow {\color{Blue} z=2\left ( \cos \left ( 0 \right ) +i\sin \left ( 0 \right )\right )}\Rightarrow {\color{Blue} z=2}\Rightarrow {\color{Blue} \left ( z-2 \right )=0}\\\\ if\ \ {\color{Blue} k=1}\Rightarrow {\color{Blue} z=2\left ( \cos \left ( \frac{2\pi }{5} \right ) +i\sin \left ( \frac{2\pi }{5} \right )\right )}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{2\pi }{5} \right )i}}\Rightarrow {\color{Blue} \left (z-2e^{\left ( \frac{2\pi }{5} \right )i} \right )=0}$

$\\ if\ \ {\color{Blue} k=2}\Rightarrow {\color{Blue} z=2\left ( \cos \left ( \frac{4\pi }{5} \right ) +i\sin \left ( \frac{4\pi }{5} \right )\right )}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{4\pi }{5} \right )i}}\Rightarrow {\color{Blue} \left (z-2e^{\left ( \frac{4\pi }{5} \right )i} \right )=0}$

$\\ if\ \ {\color{Blue} k=3}\Rightarrow {\color{Blue} z=2\left ( \cos \left ( \frac{6\pi }{5} \right ) +i\sin \left ( \frac{6\pi }{5} \right )\right )}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{6\pi }{5} \right )i}}\\ \Rightarrow {\color{Blue} z=2e^{\left (\frac{-4\pi }{5} \right )i}.e^{\left ( 2\pi \right )i}}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{-4\pi }{5} \right )i}}\Rightarrow {\color{Blue} \left ( z-2e^{\left ( \frac{-4\pi }{5} \right )i} \right )=0}$

$\\ if\ \ {\color{Blue} k=4}\Rightarrow {\color{Blue} z=2\left ( \cos \left ( \frac{8\pi }{5} \right ) +i\sin \left ( \frac{8\pi }{5} \right )\right )}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{8\pi }{5} \right )i}}\\ \Rightarrow {\color{Blue} z=2e^{\left (\frac{-2\pi }{5} \right )i}.e^{\left ( 2\pi \right )i}}\Rightarrow {\color{Blue} z=2e^{\left ( \frac{-2\pi }{5} \right )i}}\Rightarrow {\color{Blue} \left ( z-2e^{\left ( \frac{-2\pi }{5} \right )i} \right )=0}$

$\\ \therefore \ \ {\color{Blue} \left ( z-2 \right )\left ( z-2e^{\left ( \frac{2\pi }{5} \right )i} \right )\left ( z-2e^{\left ( \frac{4\pi }{5} \right )i} \right )\left ( z-2e^{\left ( \frac{-4\pi }{5} \right )i} \right )\left ( z-2e^{\left ( \frac{-2\pi }{5} \right )i} \right )=0}\\ \Rightarrow {\color{Blue} \left ( z-2 \right )}{\color{Magenta} \left ( z-2e^{\left ( \frac{2\pi }{5} \right )i} \right )\left ( z-2e^{\left ( \frac{-2\pi }{5} \right )i} \right )}{\color{DarkGreen}\left ( z-2e^{\left ( \frac{4\pi }{5} \right )i} \right )\left ( z-2e^{\left ( \frac{-4\pi }{5} \right )i} \right ) }{\color{Blue} =0}$

$\\ \Rightarrow {\color{Blue} \left ( z-2 \right )}{\color{Magenta}\left ( z^{2}-2\left ( e^{\left ( \frac{2\pi }{5} \right )i}+e^{\left ( \frac{-2\pi }{5} \right )i} \right )z +4e^{0}\right ) }{\color{DarkGreen} \left ( z^{2}-2\left ( e^{\left ( \frac{4\pi }{5} \right )i}+e^{\left ( \frac{-4\pi }{5} \right )i} \right )z +4e^{0}\right ) }{\color{Blue} =0}\\ \Rightarrow {\color{blue} \left ( z -2\right )\left ( z^{2} -4\cos \left ( \frac{2\pi }{5} \right )z+4\right )\left ( z^{2}-4\cos \left ( \frac{4\pi }{5} \right )z+4 \right )=0....\left ( 2 \right )}$

من المعادلتين (1) , (2) نستنتج أن :

$\\ {\color{red} \left ( z^{5}-32 \right )=\left ( z -2\right )\left ( z^{2} -4\cos \left ( \frac{2\pi }{5} \right )z+4\right )\left ( z^{2}-4\cos \left ( \frac{4\pi }{5} \right )z+4 \right )}$

الملف الشخصي · 11-08-2014, 05:09 PM

حل السؤال الأول :

$\\ {\color{Blue} \left ( z+1 \right )^{7}+\left ( z+2 \right )^{7}=0}\Rightarrow \frac{\left ( z+1 \right )^{7}}{-\left ( z+2 \right )^{7}}+\frac{\left ( z+2\right )^{7}}{-\left ( z+2 \right )^{7}}=0\Rightarrow \frac{\left ( z+1 \right )^{7}}{\left ( -z-2 \right )^{7}}-1=0 \\ \Rightarrow \left ( \frac{z+1}{-z-2} \right )^{7}=1=1\left ( \cos \left ( 0 \right ) +i\sin \left ( 0 \right )\right )\Rightarrow \frac{\left ( z+1 \right )}{\left ( -z-2 \right )}=\left ( 1\left ( \cos \left ( 0 \right )+i\sin \left ( 0 \right ) \right ) \right )^{\left ( \frac{1}{7} \right )}$

$\\ \Rightarrow \frac{\left ( z+1 \right )}{\left ( -z-2 \right )}=1^{\left ( \frac{1}{7} \right )}\left ( \cos \left ( \frac{0+2k\pi }{7} \right )+i\sin \left ( \frac{0+2k\pi }{7} \right ) \right ) \ \ \ \ k=0,1,2,3,4,5,6 \\ \therefore \frac{\left ( z+1 \right )}{\left ( -z-2 \right )}=1^{\left ( \frac{1}{7} \right )}e^{\left ( \frac{0+2k\pi }{7} \right )i}=e^{\left ( \frac{2k\pi }{7} \right )i}\Rightarrow \left ( z+1 \right )=e^{\left ( \frac{2k\pi }{7} \right )i}\left ( -z-2 \right ) \\ \Rightarrow z+1=-z.e^{\left ( \frac{2k\pi }{7} \right )i}-2e^{\left ( \frac{2k\pi }{7} \right )i}\Rightarrow z+z.e^{\left ( \frac{2k\pi }{7} \right )i}=-2e^{\left ( \frac{2k\pi }{7} \right )i}-1$

$\\ \Rightarrow \left [ z.{\color{Blue} e^{\left ( \frac{k\pi }{7} \right )i}}\left ( e^{\left ( \frac{-k\pi }{7} \right )i}+e^{\left ( \frac{k\pi }{7} \right )i} \right ) =\left ( -1 \right ){\color{Blue} e^{\left ( \frac{k\pi }{7} \right )i}}\left ( 2e^{\left ( \frac{k\pi }{7} \right )i}+e^{\left ( \frac{-k\pi }{7} \right )i} \right )\right ]\div {\color{Blue} e^{\left ( \frac{k\pi }{7} \right )i}} \\ \Rightarrow z\left ( e^{\left ( \frac{k\pi }{7} \right )i} +e^{\left ( \frac{-k\pi }{7} \right )i}\right )=-\left ( \left ( e^{\left ( \frac{k\pi }{7} \right )i}+e^{\left ( \frac{-k\pi }{7} \right )i} \right )+e^{\left ( \frac{k\pi }{7} \right )i} \right )$

$\\ \Rightarrow z\left ( 2\cos \left ( \frac{k\pi }{7} \right ) \right )=-\left ( \left ( 2\cos \left ( \frac{k\pi }{7} \right ) \right ) +\cos \left ( \frac{k\pi }{7} \right )+i\sin \left ( \frac{k\pi }{7} \right )\right ) \\ \\ \Rightarrow z=\frac{-3\cos \left ( \frac{k\pi }{7} \right )-i\sin \left ( \frac{k\pi }{7} \right )}{2\cos \left ( \frac{k\pi }{7} \right )}=\frac{-3\cos \left ( \frac{k\pi }{7} \right )}{2\cos \left ( \frac{k\pi }{7} \right )}-\frac{i\sin \left ( \frac{k\pi }{7} \right )}{2\cos \left ( \frac{k\pi }{7} \right )}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2} \tan \left ( \frac{k\pi }{7} \right )\right )i}$

$\\ if \ \ \ {\color{Blue} k=0}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{\left (0 \right )\pi }{7} \right ) \right )i}\Rightarrow {\color{Blue} z=\frac{-3}{2}} \\ \\ if \ \ \ {\color{Blue} k=1}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{\pi }{7} \right ) \right )i} \\ \\ if \ \ \ {\color{Blue} k=2}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{2\pi }{7} \right ) \right )i} \\ \\ if \ \ \ {\color{Blue} k=3}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{3\pi }{7} \right ) \right )i}$

$\\ if \ \ \ {\color{Blue} k=4}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{4\pi }{7} \right ) \right )i} \\ \\ if \ \ \ {\color{Blue} k=5}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{5\pi }{7} \right ) \right )i} \\ \\ if \ \ \ {\color{Blue} k=6}\Rightarrow {\color{Blue} z=\frac{-3}{2}-\left ( \frac{1}{2}\tan \left ( \frac{6\pi }{7} \right ) \right )i}$

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