1708\hspace{-16}\mathbf{x+iy=\left(1-i\sqrt{3}\right)^{100}=(-2)^{100}.\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)^{100}}$\\\\\\ As We Know that $\mathbf{-\frac{1}{2}+i\frac{\sqrt{3}}{2}=\omega}\;\;$(Cube Root of Unity.)\\\\\\ So $\mathbf{x+iy=2^{100}.\omega^{100}=2^{100}.\omega}$\\\\\\ So $\mathbf{x+iy=2^{100}.\left(-\frac{1}{2}+\frac{i.\sqrt{3}}{2}\right)=-2^{99}+i.2^{99}.\sqrt{3}}$\\\\\\ so $\boxed{\boxed{\mathbf{\left(x,y\right)=\left(-2^{99}\;, 2^{99}.\sqrt{3}\right)}}}$
