We first need to solve equation `x^2+x+1=0`. We can do this by using the following formula for solution o equation `ax^2+bx+c=0`:
`x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)`
Thus we have
`x_(1,2)=(-1) pm sqrt(1-4)/2=(-1 pm i sqrt(3))/2`
Therefore `a= ` `(-1- i sqrt(3))/2=-1/2-i sqrt(3)/2` and `b=` `(-1 + i sqrt(3))/2=-1/2+i sqrt(3)/2`
Let' now write `a` and `b` in trigonometric form:
`a` is in third quadrant therefore
`arg(a)=arctan((-sqrt(3)/2)/(-1/2))+pi=(4pi)/3`
`|a|=sqrt((1/2)^2+(sqrt(3)/2)^2)=1`
Therefore `a=cos((4pi)/3)+i sin((4pi)/3)`
Similary `b` is in second quadrant:
`arg(b)=` `arctan((sqrt(3)/2)/(-1/2))+pi=(2pi)/3`
`|b|=1`
`b=cos((2pi)/3)+i sin((2pi)/3)`
Now for power of complex number we can use de Moivre's formula:
`(cos x+ i sin x)^n=cos(nx)+i sin(nx)`
Hence
`a^1980=(cos((4pi)/3)+i sin((4pi)/3))^1980=cos(660cdot4pi)+i sin(660cdot4pi)`
`=cos0+isin0=1`
`b^1980=(cos((2pi)/3)+i sin((2pi)/3))^1980=cos(660cdot2pi)+i sin(660cdot2pi)`
`=cos0+isin0=1`
Therefore `a^1980+b^1980=1+1=2`