Найти алгебраическую форму комплексного числа

Ответ:

$(\frac{(\sqrt{2} -i*\sqrt{2} )*(1-i*\sqrt{3} )}{4} )^{28}$

=[tex]\frac{\sqrt{2}-\sqrt{2}*i-\sqrt{6}*i+\sqrt{6}*i^{2} }{4}[tex]

=[tex]\frac{\sqrt{2} -\sqrt{6} -(\sqrt{2}+\sqrt{6} ) *i}{4}" alt="\frac{(\sqrt{2} -i*\sqrt{2} )*(1-i*\sqrt{3} )}{4}[tex]

=[tex]\frac{\sqrt{2}-\sqrt{2}*i-\sqrt{6}*i+\sqrt{6}*i^{2} }{4}[tex]

=[tex]\frac{\sqrt{2} -\sqrt{6} -(\sqrt{2}+\sqrt{6} ) *i}{4}" align="absmiddle" class="latex-formula">

$Z=\frac{\sqrt{2} -\sqrt{6} -i*(\sqrt{2} +\sqrt{6} )}{4}$

r= $\sqrt{(\frac{\sqrt{2} -\sqrt{6} }{4} )^{2} +(\frac{\sqrt{2} +\sqrt{6} }{4} )^{2}} =\sqrt{\frac{2+6-2\sqrt{12}+2+6+2\sqrt{12} }{16} } =\sqrt{1}=1$

Z=r $e^{\alpha 2*i}$ где $\alpha =arctg\frac{\sqrt{2}+\sqrt{6} }{\sqrt{2}-\sqrt{6} }$ =

$arctg\frac{(\sqrt{2} +\sqrt{6} )^{2} }{(\sqrt{2}-\sqrt{6} )(\sqrt{2} +\sqrt{6} )}$ = $arctg(-2-\sqrt{3} )$

Z²⁸=1²⁸* $e^{28i*\alpha }$