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

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

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

Z²⁸=1²⁸*$e^{28i*alpha }$