Let p,q R and (1-√3 i)^200=2^199(p+iq), i=-1.

Question

Let $p,q\in\mathbb R$ and $(1-\sqrt3 i)^{200}=2^{199}(p+iq),\ i=\sqrt{-1}$. Then $p+q+q^2$ and $p-q+q^2$ are roots of the equation:

Accepted Answer

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

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

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

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

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

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

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

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

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

$$\therefore$$ 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

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

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

Solutions · 1