Question

For any complex number $w=c+$ id, let $\arg (w)\in (-\pi ,\pi ]$, where $i=\sqrt{-1}$. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers $z=x+$ iy satisfying $\arg \left(\frac{z+\alpha }{z+\beta }\right)=\frac{\pi }{4}$, then ordered pair $(x,y)$ lies on the circle $x^2+y^2+5x-3y+4=0$. Then which of the following statements is(are) TRUE?

• A. $\alpha =-1$
• B. $\alpha \beta =4$
• C. $\alpha \beta =-4$
• D. $\beta =4$

Answer 1

Answer: (B), (D)

$w=c+id,\arg (w)\in (-\pi ,\pi ]$
$\arg \left(\frac{z+\alpha }{z+\beta }\right)=\frac{\pi }{4};z_0=-\frac{5}{2}+\frac{3}{2}i$
$z_0+\alpha =\left(z_0+\beta \right)i;z_0+\alpha =z_{0}i+\beta i;z_0(1-i)=\beta i-\alpha$ (by Rotation)
$-\frac{5}{2}+\frac{3}{2}i+\alpha =\left(-\frac{5}{2}+\frac{3}{2}i\right)i+\beta i$
$-\frac{5}{2}+\frac{3}{2}i+\alpha =-\frac{5}{2}i-\frac{3}{2}+\beta i$
$-\frac{5}{2}+\frac{3}{2}+\frac{3}{2}i+\frac{5}{2}i+\alpha =\beta i;-1+4i=\beta i-\alpha ($ As $\alpha ,\beta$ are real number $)$
$\Rightarrow -\alpha =-1,\alpha =1;\beta =4$