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 $x^{2}+y^{2}+5 x-3 y+4=0$. 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)

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