Question

If a, b, g, d are four complex numbers such that $\frac{\gamma}{\delta}$is real and ad – bg ¹ 0, then z = $\frac{\alpha + \beta t}{\gamma + \delta t}$, tĪ R represents a

• A. Circle
• B. Parabola
• C. Ellipse
• D. Straight line

Answer 1

Answer: (D)

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 ($\bar{\gamma}$d – g$\bar{\delta}$)z$\bar{z}$+(g$\bar{\beta}$–$\bar{\alpha}$d)z + (a$\bar{\delta}$ – b$\bar{\gamma}$)$\bar{z}$

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

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

Therefore (1) can be written as $\bar{a}$z + a$\bar{z}$ = c ...(2)

where a = i(a$\bar{\delta}$ – b$\bar{\gamma}$) and c = i($\bar{\alpha}$b – a$\bar{\beta}$)

Note that a ¹ 0 for if a = 0 then

a$\bar{\delta}$ – b$\bar{\gamma}$ = 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 hypothesis.

Also, note that c = i($\bar{\alpha}$b – a$\bar{\beta}$) is a purely real number.

Thus, z = $\frac{\alpha + \beta t}{\gamma + \delta t}$ represents a straight line.