تحويل للشكل المثلثي

الملف الشخصي · 10-30-2014, 05:13 PM

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الملف الشخصي · 11-01-2014, 04:44 PM

${\color{Blue} z=1-\sqrt{2}+i}$

${\color{Blue} r=\sqrt{x^{2}+y^{2}}}\Rightarrow r=\sqrt{\left ( 1-\sqrt{2} \right )^{2}+1^{2}}=\sqrt{1-2\sqrt{2}+2+1}=\sqrt{4-2\sqrt{2}}$

${\color{Blue} \cos \left ( \Theta \right )=\frac{x}{r}}\Rightarrow \cos \left ( \Theta \right )=\frac{1-\sqrt{2}}{\sqrt{4-2\sqrt{2}}}< 0$

${\color{Blue} \sin \left ( \Theta \right )=\frac{y}{r}}\Rightarrow \sin \left ( \Theta \right )=\frac{1}{\sqrt{4-2\sqrt{2}}}> 0$

$\therefore {\color{Blue} \Theta \in \left ( \frac{\pi }{2},\pi \right )}$

$\therefore {\color{Blue} z=r\left ( \cos \left ( \Theta \right )+i\sin \left ( \Theta \right ) \right ),\Theta \in \left ( \frac{\pi }{2},\pi \right )}$

$\Rightarrow z=\sqrt{4-2\sqrt{2}}\left ( \cos \left ( \Theta \right ) +i\sin \left ( \Theta \right ) \right ),\Theta \in \left ( \frac{\pi }{2},\pi \right )$

$\therefore {\color{Blue} z^{2}=\left ( \sqrt{4-2\sqrt{2}} \left ( \cos \left ( \Theta \right )+i\sin \left ( \Theta \right ) \right )\right )^{2}}$

$\Rightarrow z^{2}=\left ( 4-2\sqrt{2} \right )\left ( \cos \left ( 2\Theta \right )+i\sin \left ( 2\Theta \right ) \right ),{\color{Blue} 2\Theta \in \left ( \pi ,2\pi \right )}$

$\tan \left ( \Theta \right )=\frac{\sin \left ( \Theta \right )}{\cos \left ( \Theta \right )}=\frac{\left ( \frac{1}{\sqrt{4-2\sqrt{2}}} \right )}{\left ( \frac{1-\sqrt{2}}{\sqrt{4-2\sqrt{2}}} \right )}=\frac{1}{\left ( 1-\sqrt{2} \right )}$

${\color{Blue} \tan \left ( 2\Theta \right )=\frac{2\tan \left ( \Theta \right )}{1-\tan ^{2}\left ( \Theta \right )}=\frac{2\left ( \frac{1}{1-\sqrt{2}} \right )}{1-\left ( \frac{1}{1-\sqrt{2}} \right )^{2}}\left ( \frac{\left (1-\sqrt{2} \right )^{2}}{\left (1-\sqrt{2} \right )^{2}} \right )=\frac{2\left ( 1-\sqrt{2} \right )}{\left ( 1-\sqrt{2} \right )^{2}-1}}$

$\Rightarrow {\color{Blue}\tan \left ( 2\Theta \right )= \frac{2-2\sqrt{2}}{1-2\sqrt{2}+2-1}=\frac{\left ( 2-2\sqrt{2} \right )}{\left (2-2\sqrt{2} \right )}=1> 0}$

$\because {\color{Blue} \left ( \tan \left ( 2\Theta \right ) =1\right )}and{\color{Blue} \left ( 2\Theta \in \left ( \pi ,2\pi \right ) \right )}\Rightarrow {\color{Blue} 2\Theta =\pi +\frac{\pi }{4}=\frac{5\pi }{4}}\Rightarrow {\color{Blue} \Theta =\frac{5\pi }{8}\in \left ( \frac{\pi }{2},\pi \right )}$

$\therefore {\color{Blue} z=\sqrt{4-2\sqrt{2}}\left ( \cos \left ( \frac{5\pi }{8} \right ) +i\sin \left ( \frac{5\pi }{8} \right )\right )}$

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تحويل للشكل المثلثي

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