If (√3+i)100=299(p+i q), then p and q are roots of the equation :

Question

If (√3+i)100=299(p+i q), then p and q are roots of the equation : (A) x2-(√3-1) x-√3=0 (B) x2+(√3+1) x+√3=0 (C) x2+(√3-1) x-√3=0 (D) x2-(√3+1) x+√3=0

Tardigrade Admin · Accepted Answer

(2 e i π / 6)100=299( p + iq ) 2100( cos (50 π/3)+ i sin (50 π/3)) =299( p + iq ) p + iq =2( cos (2 π/3)+ i sin (2 π/3)) p =-1, q =√3 x 2-(√3-1) x -√3=0

Q. If $(3 + i)^{100} = 2^{99} (p + i q)$ , then $p$ and $q$ are roots of the equation :

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

$x^{2} + (3 + 1) x + 3 = 0$

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

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

$(2 e^{iπ /6})^{100} = 2^{99} (p + i q)$
$2^{100} (cos \frac{50 π}{3} + i sin \frac{50 π}{3})$
$= 2^{99} (p + i q)$
$p + i q = 2 (cos \frac{2 π}{3} + i sin \frac{2 π}{3})$
$p = - 1,$
$q = 3$
$x^{2} - (3 - 1) x - 3 = 0$

Q. If (3​+i)100=299(p+iq), then p and q are roots of the equation :

x2−(3​−1)x−3​=0

x2+(3​+1)x+3​=0

x2+(3​−1)x−3​=0

x2−(3​+1)x+3​=0