If ${\left( {\sqrt 3 + i} \right)^{100}} = {2^{99}}(p + iq)$, then p and q are roots of the equation :
Solution
${\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)$<br><br>$${2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)$$<br><br>$p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$<br><br>p = $-$1, q = $\sqrt 3$<br><br>${x^2} - (\sqrt 3 - 1)x - \sqrt 3 = 0$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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