Корень ис чесла 9а-4b,если а=16,b=11

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= (cos(-π/12)+isin(-π/12))= (cos(π/12)-isin(π/12))=

=((√6+√2)/4-i(√6-√2)/4)=((√6+√2)-i(√6-√2))/4

x₁= (cos((-π+2π)/12)+isin((-π+2π)/12))= (cos((π)/12)+isin((π)/12))=

((√6+√2)+i(√6-√2))/4

x₂= (cos((-π+4π)/12)+isin((-π+4π)/12))= (cos((3π)/12)+isin((3π)/12))=

= (cos((π)/4)+isin((π)/4))= (√2/2+i√2/2)= √2(1+i)/2

x₃= (cos((-π+6π)/12)+isin((-π+6π)/12))= (cos((5π)/12)+isin((5π)/12))=

= (sin((π)/12)+icos((π)/12))= ((√6-√2)/4+i(√6+√2)/4)=

=((√6-√2)+i(√6+√2))/4

x₄= (cos((-π+8π)/12)+isin((-π+8π)/12))= (cos((7π)/12)+isin((7π)/12))=

= (-sin((π)/12)+icos((π)/12))= (-(√6-√2)/4+i(√6+√2)/4)=

=((√2-√6)+i(√6+√2))/4

x₅= (cos((-π+10π)/12)+isin((-π+10π)/12))= (cos((3π)/4)+isin((3π)/4))=

= (-cos((π)/4)+isin((π)/4))= (-√2/2+i√2/2)= √2(-1+i)/2