Question

lf $(x+iy)(2+cos\theta +isin\theta )=3$ then $x^2+y^2-4x+3$ is

• A. $0$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A $0$

$writingas\theta =cos\theta +isin\theta$
$2(x+iy)+(x+iy)(cos\theta +isin\theta )=3$
$\Rightarrow (2x+xcos\theta -ysin\theta )+i(2y+xsin\theta +ycos\theta )=3$
then by comparison,
$2x+xcos\theta -ysin\theta =3orxcos\theta -ysin\theta =3-2x-\left(1\right)$
$2y+xsin\theta +ycos\theta =0xsin\theta +ycos\theta =-2y-\left(2\right)$
By (1) & (2), squaring and adding,
$x^2(co{s}^2\theta +si{n}^2\theta )+y^2(si{n}^2\theta +co{s}^2\theta )$
$-2xcos\theta ysin\theta +2xcos\theta ysin\theta =(3-2x{)}^2+(-2y{)}^2$
$\Rightarrow x^2+y^2=a+4x^2-12x+4y^2$
$\Rightarrow 3x^2+3y^2-12x+9=0$
$\Rightarrow x^2+y^2-4x+3=0$