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.