Question

Let $z$ be a complex number such that $\arg \left(\frac{z-2}{2z+3i}\right)=0$, then the locus of $z$ is

• A. a straight line with slope $\frac{4}{3}$
• B. a pair of straight lines passing through the origin
• C. a straight line with slope $\frac{3}{4}$
• D. a pair of straight lines passing through $(1,1)$

Answer 1

The correct option is C a straight line with slope $\frac{3}{4}$

Let $z=x+iy$
Now, $\frac{z-2}{2z+3i}$
$=\frac{(x-2)+iy}{2x+i(2y+3)}$
$=\frac{(x-2)+iy}{2x+i(2y+3)}\times \frac{2x-i(2y+3)}{2x-i(2y+3)}$
$=\frac{(x-2)2x-i(x-2)(2y+3)+i2xy+y(2y+3)}{4x^2+(2y+3{)}^2}$

Since, $\arg \left(\frac{z-2}{2z+3i}\right)=0$
$\Rightarrow \frac{z-2}{2z+3i}$ is purely real.
$\Rightarrow 2xy-(x-2)(2y+3)=0$
$\Rightarrow 3x-4y-6=0$
which represents a straight line with slope $\frac{3}{4}.$

Alternate:
$\arg \left(\frac{z-\alpha }{z-\beta }\right)=\theta$ represents the arc of the circle, where the chord joining $P\left(\alpha \right)$ and $Q\left(\beta \right)$ subtends angle $\theta$ to the circumference.


Here, as the angle is $0$, the locus of the curve will be a straight line passing through $P\left(\alpha \right)$ and $Q\left(\beta \right)$
$P\left(\alpha \right)=(2,0)$ and $Q\left(\beta \right)=(0,\frac{-3}{2})$
The straight line passing through $(2,0)$ and $(0,\frac{-3}{2})$ is $3x-4y-6=0$