If (√3+i)^100=2^99(p+iq), then p and q are roots of the equation

Question

If $(\sqrt3+i)^{100}=2^{99}(p+iq)$, then $p$ and $q$ are roots of the equation

Accepted Answer

$${\left( {2{e^{i\pi /6}}} \right)^{100}} = {2^{99}}(p + iq)$$

$${2^{100}}\left( {\cos {{50\pi } \over 3} + i\sin {{50\pi } \over 3}} \right) = {2^{99}}(p + iq)$$

$$p + iq = 2\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)$$

p = -1, q = $$\sqrt 3 $$

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

Solutions · 1