Question

помогите пожалуйста решить уравнения в поле комплексных чисел
$x {}^{6} + 3 i = 0$
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Answer 1

Ответ:

Объяснение:

Уравнение к-ой степени имеет к корней в поле комплексных чисел

cos(π/12)=cos[(π/3)-(π/4)]=cos(π/3)cos(π/4)+sin(π/3)sin(π/4)=

=0,5·(√2/2)+(√3/2)(√2/2)=(√6+√2)/4

sin(π/12)=sin[(π/3)-(π/4)]=sin(π/3)cos(π/4)-cos(π/3)sin(π/4)=

=(√3/2)(√2/2)-0,5·(√2/2)=(√6-√2)/4

x⁶+3i=0

x⁶=-3i=3(cos(-π/2)+isin(-π/2))

x₀=$\sqrt[6]{3}$ (cos(-π/12)+isin(-π/12))=$\sqrt[6]{3}$ (cos(π/12)-isin(π/12))=

=$\sqrt[6]{3}$((√6+√2)/4-i(√6-√2)/4)=$\sqrt[6]{3}$((√6+√2)-i(√6-√2))/4

x₁=$\sqrt[6]{3}$ (cos((-π+2π)/12)+isin((-π+2π)/12))=$\sqrt[6]{3}$ (cos((π)/12)+isin((π)/12))=

$\sqrt[6]{3}$((√6+√2)+i(√6-√2))/4

x₂=$\sqrt[6]{3}$ (cos((-π+4π)/12)+isin((-π+4π)/12))=$\sqrt[6]{3}$ (cos((3π)/12)+isin((3π)/12))=

=$\sqrt[6]{3}$ (cos((π)/4)+isin((π)/4))=$\sqrt[6]{3}$ (√2/2+i√2/2)=$\sqrt[6]{3}$ √2(1+i)/2

x₃=$\sqrt[6]{3}$ (cos((-π+6π)/12)+isin((-π+6π)/12))=$\sqrt[6]{3}$ (cos((5π)/12)+isin((5π)/12))=

=$\sqrt[6]{3}$ (sin((π)/12)+icos((π)/12))=$\sqrt[6]{3}$ ((√6-√2)/4+i(√6+√2)/4)=

=$\sqrt[6]{3}$((√6-√2)+i(√6+√2))/4

x₄=$\sqrt[6]{3}$ (cos((-π+8π)/12)+isin((-π+8π)/12))=$\sqrt[6]{3}$ (cos((7π)/12)+isin((7π)/12))=

=$\sqrt[6]{3}$ (-sin((π)/12)+icos((π)/12))=$\sqrt[6]{3}$ (-(√6-√2)/4+i(√6+√2)/4)=

=$\sqrt[6]{3}$((√2-√6)+i(√6+√2))/4

x₅=$\sqrt[6]{3}$ (cos((-π+10π)/12)+isin((-π+10π)/12))=$\sqrt[6]{3}$ (cos((3π)/4)+isin((3π)/4))=

=$\sqrt[6]{3}$ (-cos((π)/4)+isin((π)/4))=$\sqrt[6]{3}$ (-√2/2+i√2/2)=$\sqrt[6]{3}$ √2(-1+i)/2