Question

If a, b, g, d are four complex numbers such that g/d is real and ad – bg¹ 0, then z = $\frac{\alpha + \beta t}{\gamma + \delta t}$, t Ī R represents –

• A. Circle
• B. Parabola
• C. Straight line
• D. None of these

Answer 1

Answer: (C)

Straight line

Answer 2

Sol. z = $\frac{\alpha + \beta t}{\gamma + \delta t}$Ž (g + dt) z = a + bt Ž (dz – b)t = a – gz

Ž t = $\frac{\alpha - \gamma z}{\delta z - \beta}$ [Q ad – bg ¹ 0]

As t is real, $\frac{\alpha - \gamma z}{\delta z - \beta}$ = $\frac{\bar{\alpha} - \bar{\gamma}\bar{z}}{\bar{\delta}\bar{z}–\bar{\beta}}$

Ž (a – gz) $(\bar{\delta}\bar{z}–\bar{\beta})$ = $(\bar{\alpha}–\bar{\gamma}\bar{z})$ (dz – b)

Ž $(\bar{\gamma}\delta - \gamma\bar{\delta})$ z$\bar{z}$ + $(\gamma\bar{\beta} - \bar{\alpha}\delta)$ z + ($\alpha\bar{\delta}–\beta\bar{\gamma}$)$\bar{z}$

= a$\bar{\beta}$ – $\bar{\alpha}$b … (1)

Since $\frac{\gamma}{\delta}$ is real, $\frac{\gamma}{\delta}$= $\frac{\bar{\gamma}}{\bar{\delta}}$or g$\bar{\delta}$ – $\bar{\gamma}$d = 0.

Therefore (1) can be written as

$\bar{a}z + a\bar{z} = c$ … (2)

Where a = –I ($\alpha\bar{\delta} - \beta\bar{\gamma}$) and c = i ($\bar{\alpha}\beta - \alpha\bar{\beta}$)

Note that a ¹ 0 for if a = 0 then $\alpha\bar{\delta} - \beta\bar{\gamma}$ = 0

Ž $\frac{\alpha}{\beta} = \frac{\bar{\gamma}}{\bar{\delta}} = \frac{\gamma}{\delta}$ (Q $\frac{\gamma}{\delta}$ is real)

Ž ad – bg = 0.

Which is against the hypothesis. Also not that (z) is a straight line not a circle.