### Instruction: Evaluate $(2-w)(2-w^2)\cdots(2-w^{10})$ where $w=e^{2\pi i/11}.$.### Response: I notice that $w$ is a complex root of unity, so $w^{11}=1$ and $w$ satisfies the equation $x^{11}-1=0.$
I also notice that the factors in the product are of the form $2-w^k$ for $k=0,1,\ldots,10.$
This makes me think of the factorization of $x^{11}-1$ into $(x-1)(x-w)(x-w^2)\cdots(x-w^{10}).$
If I plug in $x=2$ into this factorization, I get $2^{11}-1=(2-1)(2-w)(2-w^2)\cdots(2-w^{10}).$
This means that $(2-w)(2-w^2)\cdots(2-w^{10})=\frac{2^{11}-1}{2-1}=\frac{2047}{1}=2047.$