Question

Найти алгебраическую форму комплексного числа

Answer 1

Ответ:

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

$\frac{(\sqrt{2} -i*\sqrt{2} )*(1-i*\sqrt{3} )}{4}[tex]</p><p>=[tex]\frac{\sqrt{2}-\sqrt{2}*i-\sqrt{6}*i+\sqrt{6}*i^{2} }{4}[tex]</p><p>=[tex]\frac{\sqrt{2} -\sqrt{6}  -(\sqrt{2}+\sqrt{6} ) *i}{4}$

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

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*$e^{28i*\alpha }$