sum_03 (roots $^{200}$ )

If $f(x) = 0$ , then

$\begin{aligned} 1 + x + x^2 + \cdots + x^{199} & = 0 & \small \color{#3D99F6} \text{Multiply both sides by }x \\ x+x^2 + x^3 + \cdots + x^{200} & = 0 & \small \color{#3D99F6} \text{Add both sides by }1 \\ {\color{#3D99F6}1+ x + x^2 + \cdots + x^{199}} + x^{200} & = 1 & \small \color{#3D99F6} \text{Note that }1 + x + x^2 + \cdots + x^{199} = 0 \\ \implies x^{200} & = 1 \end{aligned}$

Since $a_k$ , where $k=1, 2, 3, \cdots 199$ , are roots of $f(x) = 0$ . Then $a_k^{200} = 1$ and $\displaystyle \sum_{k=1}^{199} a_k^{200} = \sum_{k=1}^{199} 1 = \boxed{199}$ .

$1+x+x^2+x^3+...+x^{198}+x^{199}=0$ ,so $(1+x+x^2+x^3+...+x^{198}+x^{199})(x-1)=0=x^{200}-1,x^{200}=1$

So, $a_1^{200}=a_2^{200}=a_3^{200}=...=a_{199}^{200}=1$ .Hence ,the answer is 199.

By definition, $1 + a_m + a_m^2 + \cdots + a_m^{199} = 0$ for $m=1,2,\ldots,199$ . Then,

$\sum_{k=1}^{199} a_k^{200} = \sum_{k=1}^{199} (a_k^{200} - 1 + 1)= \sum_{k=1}^{199} (a_k - 1)\cancelto0{(1 + a_k + a_k^2 + \cdots + a_k^{199} )} + \sum_{k=1}^{199} 1 = 0 + 199 = \boxed{199} .$

$\begin{aligned} &f(x)=1+x+x^2+x^3+ \cdots +x^{199} = \frac{x^{200}-1}{x-1}\\ &\text{let }x=e^{i \theta}\\ &f(x)=0\Leftrightarrow \frac{x^{200}-1}{x-1}=0\Leftrightarrow x^{200}-1=0\\ &e^{200 i\theta}=1\Leftrightarrow \theta=\frac{2k \pi}{200} \text{ for } k=1,2,3,\cdots,199 & x\neq1\Rightarrow k\neq0\\ &a_k=e^{\frac{2k \pi i}{200}}\Rightarrow a^{200}_k=e^{2k \pi i}\\ &\sum_{k=1}^{199}{(a_k)^{200}}=\sum_{k=1}^{199}{e^{2k \pi i}}= \color{#D61F06}\sum_{k=1}^{199}{\left ( e^{2\pi i} \right )^k}=\frac{\left ( e^{2\pi i} \right )^{200}-e^{2\pi i}}{e^{2\pi i}-1} & \color{#D61F06}\text{geometric progression}\\ &\lim_{x\rightarrow 2\pi i}\frac{\left ( e^x \right )^{200}-e^{x}}{e^{x}-1}=\lim_{x\rightarrow 2\pi i}\frac{200 \left ( e^x \right )^{200}-e^{x}}{e^{x}}=\lim_{x\rightarrow 2\pi i}200e^{199x}-1=199 & \text{l'hospital rule} \end{aligned}$

Used objective approach