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...

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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`