Question

Exercițiul 1 va rog frumos am nevoie rapid de el!!!!

Answer 1

$\displaystyle \Bigg(\frac{1+i\sqrt{3} }{\sqrt{3} -i}\Bigg)^6\cdot\Big(1-i\sqrt{3}\Big)^9 \\\\ Formule:\\\\Z=a+bi=r(cos \alpha +isin\alpha);~r=\sqrt{a^2+b^2}\\\\ \frac{Z_1}{Z_2} =\frac{r_1}{r_2} [cos(\alpha_1-\alpha_2)+isin(\alpha_1-\alpha_2)]\\\\ (cos\alpha+isin\alpha)^n=cosn\alpha+isinn\alpha\\.........................................................................................$

$\displaystyle Z_1=1+i\sqrt{3} ;~r_1=\sqrt{1^2+\Big(\sqrt{3} \Big)^2} =\sqrt{1+3} =\sqrt{4} =2\\\\ Z_1=1+i\sqrt{3} =2\Bigg(\frac{1}{2}+i\frac{\sqrt{3} }{2}\Bigg) =2\Bigg(cos\frac{\pi}{3} +isin\frac{\pi}{3} \Bigg)$

$\displaystyle Z_2=\sqrt{3} -i,~r_2=\sqrt{\Big(\sqrt{3}\Big)^2+(-1)^2 } =\sqrt{3+1} =\sqrt{4} =2\\\\ Z_2=\sqrt{3} -i=2\Bigg(\frac{\sqrt{3} }{2} -i\frac{1}{2}\Bigg) =2\Bigg(cos\frac{11\pi}{6} +isin\frac{11\pi}{6}\Bigg)$

$\displaystyle Z_3=1-i\sqrt{3} ;~r_3=\sqrt{1^2+\Big(-\sqrt{3} \Big)^2} =\sqrt{1+3} =\sqrt{4} =2\\\\ Z_3=1-i\sqrt{3} =2\Bigg(\frac{1}{2} -i\frac{\sqrt{3} }{2}\Bigg) =2\Bigg(cos\frac{5\pi}{3} +i sin \frac{5\pi}{3} \Bigg)$

$\displaystyle \Bigg(\frac{1+i\sqrt{3} }{\sqrt{3} -i}\Bigg)^6\cdot\Big(1-i\sqrt{3}\Big)^9 =\\\\=\Bigg(\frac{2\Big(cos\frac{\pi}{3}+isin\frac{\pi}{3} \Big) }{2\Big(cos\frac{11\pi}{6} +isin\frac{11\pi}{6} \Big)} \Bigg)^6\cdot \Bigg[2 \Bigg(cos\frac{5\pi}{3} +isin\frac{5\pi}{3}\Bigg)\Bigg]^9=$

$\displaystyle \Bigg(\frac{cos\frac{\pi}{3}+isin\frac{\pi}{3} }{cos\frac{11\pi}{6} +isin\frac{11\pi}{6} } \Bigg)^6\cdot \Bigg[2 \Bigg(cos\frac{5\pi}{3} +isin\frac{5\pi}{3}\Bigg)\Bigg]^9=\\\\=\frac{cos\not\not6\cfrac{\pi}{\not3} +isin\not6\cfrac{\pi}{\not3} }{cos\not6\cfrac{11\pi}{\not6}+isin\not6\cfrac{11\pi}{\not6} } \cdot 2^9\Bigg(cos\not9\frac{5\pi}{\not3} +isin\not9\frac{5\pi}{\not3} \Bigg)=\\\\=\frac{cos2\pi+isin2\pi}{cos11\pi+isin11\pi} \cdot 2^9(cos15\pi+isin15\pi)=$

$\displaystyle =\frac{2}{2} [cos(2\pi-11\pi)+isin(2\pi-11\pi)] \cdot 2^9(cos15\pi+isin15\pi)=\\\\=[cos(-9\pi)+isin(-9\pi)] \cdot 2^9(cos15\pi+isin15\pi)=\\\\=(-1+0) \cdot 2^9\cdot (-1+0)=(-1) \cdot 2^9 \cdot (-1)=(-1) \cdot 512 \cdot (-1)=512$