Question

In the argand plane, the locus of $z\ne 1$ such that $Arg\left[\frac{6z^2-15z+9}{6z^2-2z-4}\right]=\frac{2\pi }{3}$, is

• A. the straight line joining the points $z=\frac{3}{2},z=\frac{-3}{2}$
• B. the straight line joining the points $z=\frac{-3}{2},z=\frac{2}{3}$
• C. segment of a circle passing through $z=\frac{3}{2},z=\frac{-2}{3}$
• D. segment of a circle passing through $z=\frac{-3}{2},z=\frac{2}{3}$

Answer 1

The correct option is C segment of a circle passing through $z=\frac{3}{2},z=\frac{-2}{3}$

$Arg\left[\frac{6z^2-15z+9}{6z^2-2z-4}\right]=Arg\left[\frac{6z^2-6z-9z+9}{6z^2-6z+4z-4}\right]$
$=Arg\left[\frac{(6z-9)(z-1)}{(6z+4)(z-1)}\right]$ $=Arg\left\{\frac{6[z-\frac{9}{6}]}{6[z+\frac{4}{6}]}\right\}$
$=Arg\left\{\frac{z-\frac{3}{2}}{z+\frac{2}{3}}\right\}$
$\therefore$ The equation represents segment of circle passing through $\frac{3}{2}$ & $\frac{-2}{3}$

Hence, option C.