If $1,{\mathrm{\alpha }}_1,{\mathrm{\alpha }}_2,\dots .{\mathrm{\alpha }}_{99}$ are roots of ${\mathrm{Z}}^{100}=1\text{ then }\sum _{1\le \mathrm{i}<\mathrm{j}\le 99} {\mathrm{\alpha }}_{\mathrm{i}}{\mathrm{\alpha }}_{\mathrm{j}}$ equal to

${\mathrm{z}}^{100}-1=(\mathrm{z}-1)\left(\mathrm{z}-{\mathrm{\alpha }}_1\right)\left(\mathrm{z}-{\mathrm{\alpha }}_2\right)\dots .\left(\mathrm{z}-{\mathrm{\alpha }}_{99}\right)=0$
Sum of roots $=1+{\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+\dots \dots +{\mathrm{\alpha }}_{99}=0$
Sum of roots taken two at a time = 0
$$\begin{gathered} \Rightarrow \left(\begin{array}{c}{\mathrm{\Sigma \alpha }}_{\mathrm{i}}{\mathrm{\alpha }}_{\mathrm{j}}\\ 1\le \mathrm{i}<j\le 99\end{array}\right)+\left({\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+\dots \dots ..{\mathrm{\alpha }}_{99}\right)=0 \\ \Rightarrow \left(\begin{array}{c}{\mathrm{\Sigma \alpha }}_{\mathrm{i}}{\mathrm{\alpha }}_{\mathrm{j}}\\ 1\le \mathrm{i}<j\le 99\end{array}\right)-1=0 \\ \Rightarrow {\mathrm{\Sigma \alpha }}_{\mathrm{i}}{\mathrm{\alpha }}_{\mathrm{j}}=1 \end{gathered}$$