Question 1080584
{{{r=sqrt(1^2+(-1)^2)=sqrt(1+1)=sqrt(2)}}}
and
{{{theta=tan^(-1)(-1/1)=(7/4)pi}}}
So then for the cube root, the modulus would be
{{{root(3,sqrt(2))=root(6,2)}}}
For the angle,
{{{A[1]=(1/3)(7/4)pi=(7/12)pi}}}
{{{A[2]=(1/3)(7/4)pi+1*(2pi)/3=(5/4)pi}}}
{{{A[3]=(1/3)(7/4)pi+2*(2pi)/3=(23/12)pi}}}
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{{{z[1]=root(6,2)*(cos((7/12)pi)+i*sin((7/12)pi))}}}
{{{z[2]=root(6,2)*(cos((5/4)pi)+i*sin((5/4)pi))}}}
{{{z[3]=root(6,2)*(cos((23/12)pi)+i*sin((23/12)pi))}}}