If $${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$$, then p and q are roots of the equation :
A
$${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$$
B
$${x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$$
C
$${x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$$
D
$${x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$$
Explanation
$${\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)$$
$${2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)$$
$$p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$$
p = $$-$$1, q = $$\sqrt 3 $$
$${x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0$$
