ИМ
Ирина Маронова
i=cos(PI/2)+i sin(PI/2),
i^(1/12)=cos((PI/2+2n*PI)/12)+i sin((PI/2+2n*PI)/12), n=0,1,2...,11
korenj(i)=(a+bi)
i^(1/12)=(a+bi)
r=sqrt(2)
z^n=r^n*e^(in fi)=r^n*(cos n fi+ isin n fi)
fi=arctg1=Pi/4
i^(1/12)=2^(1/24)*(cos(Pi/48)+isin(Pi/48)