9. Degree 99 polynomial

What happens if you multiply the expression with $w - 1$.

Let's recall the identity for the difference of two powers:

$$ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1}). $$

Therefore, for $w^n - 1^n$ we will have

$$ w^n - 1 = (w - 1)(w^{n-1} + w^{n-2} + \cdots + w + 1). $$

Let's say $w=\exp(i2\pi/n)$. For $n = 100$, the second set of parentheses contains the expression we are looking for. Now, notice that $w^n = w^{i2\pi} = 1$. This means that the left hand side of the above expression is zero. We also know that $w - 1 \not= 0$. This means that our degree-99 polynomial is equal to zero.

$$ 1 + w + w^2 + \cdots + w^{99} = 0. $$