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If $${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$$, then p and q are roots of the equation :

JEE · Math · previous-year question

  1. A.$${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$$correct
  2. B.$${x^2} + \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$$
  3. C.$${x^2} + \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$$
  4. D.$${x^2} - \left( {\sqrt 3 + 1} \right)x + \sqrt 3 = 0$$

Answer

A. $${x^2} - \left( {\sqrt 3 - 1} \right)x - \sqrt 3 = 0$$

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 = $$\sqrt 3 $$ $${x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0$$

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