# Question 4 [JEE adv, 2021]

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 $$arg\Big[\dfrac{z + \alpha}{z + \beta}\Big]=\dfrac{\pi}{4}$$, 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?

1. $$\alpha = -1$$
2. $$\alpha\beta = 4$$
3. $$\alpha\beta = -4$$
4. $$\beta = 4$$

## Concepts

Simplify the arg part.

## Solution

#### Step 1

$\begin{eqnarray} \frac{z + \alpha}{z + \beta} &=& \frac{(x + \alpha) + iy}{(x + \beta) + iy}\\ &=& \frac{(x + \alpha)(x + \beta) + y^2 + i(\beta - \alpha)y}{(x + \beta)^2 + y^2}\\ \end{eqnarray}$ $\begin{eqnarray} \implies \frac{(\beta - \alpha)y}{(x + \alpha)(x + \beta) + y^2} = \tan\frac{\pi}{4}&=&1\\ \implies x^2 + y^2 + (\alpha + \beta)y + (\alpha - \beta)y + \alpha\beta &=& 0 \end{eqnarray}$

#### Step 2

The above curve is the locus of $$z = x + iy$$ or the pair $$(x,y)$$. This must be equal to the equation of the given circle. All we need to do now is to equate the coefficients of each term.

#### Step 3

This gives the following relations $\begin{eqnarray} \alpha + \beta &=& 5\\ \alpha - \beta &=& -3\\ \alpha \beta &=& 4\\ \implies \alpha &=& 1\\ \beta &=& 4 \end{eqnarray}$ Therefore, options 1 and 4 are correct.