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 $\arg\left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$, the 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?

(B) $\alpha\beta=4$; (D) $\beta=4$

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Complex Numbers

For any complex number

w = c + i d

, let

ar g (w) \in (- π, π]

, where

i = - 1

. Let

α

and

β

be real numbers such that for all complex numbers

z = x + i y

satisfying

ar g (\frac{z + α}{z + β}) = \frac{π}{4}

, the 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?

Answer: (B), (D)

The condition

ar g (\frac{z + α}{z + β}) = \frac{π}{4}

means that z lies on an arc where the points

(- α, 0)

and

(- β, 0)

subtend an angle of

\frac{π}{4}

. The circle

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

intersects the x-axis (y=0) at points satisfying

x^{2} + 5 x + 4 = 0

, giving

x = - 1

and

x = - 4

. These are the endpoints of the chord on the x-axis. From

ar g (\frac{z + α}{z + β}) = \frac{π}{4}

, we can write

z + α = (z + β) \cdot r \cdot e^{iπ /4}

. Comparing with the geometry, we have

z + β = z + 4

giving

β = 4

, and

z + α = z - 1

giving

α = - 1

. Therefore,

α β = (- 1) (4) = 4

(note: the source solution has

α = 1

, giving

α β = 4

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · One or more correct options