$${\left( {1 - \sqrt 3 i} \right)^{200}}$$

$$ = {\left[ {2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)} \right]^{200}}$$

$$ = {2^{200}}{\left( {\cos {\pi \over 3} - i\sin {\pi \over 3}} \right)^{200}}$$

$$ = {2^{200}}\left( {\cos {{200\pi } \over 3} - i\sin {{200\pi } \over 3}} \right)$$

$$ = {2^{200}}\left( {\cos \left( {66\pi + {{2\pi } \over 3}} \right) - i\sin \left( {66\theta + {{2\pi } \over 3}} \right)} \right)$$

$$ = {2^{200}}\left( {\cos {{2\pi } \over 3} - i\sin {{2\pi } \over 3}} \right)$$

$$ = {2^{200}}\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right)$$

$$ = {2^{199}}\left( { - 1 - \sqrt 3 i} \right)$$

$$ = {2^{199}}\left( {p + iq} \right)$$

$$\therefore$$ p = $$-$$1 and q = $$- \sqrt3$$

Now, $$p - q + {q^2} = - 1 + \sqrt 3 + 3 = 2 + \sqrt 3 = \alpha $$

and $$p + q + {q^2} = - 1 - \sqrt 3 + 3 = 2 - \sqrt 3 = \beta $$

$$\therefore$$ $$\alpha + \beta = 4$$

$$\alpha \beta = \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 4 - 3 = 1$$

$$\therefore$$ Quadratic equation is

$${x^2} - (\alpha + \beta )x + \alpha \beta = 0$$

$$ \Rightarrow {x^2} - 4x + 1 = 0$$