Question

If the locus of the complex number $z$ given by $arg\left(z+i\right)-arg\left(z-i\right)=\frac{2\pi }{3}$ is an arc of a circle, then the length of the arc is

• A. $\frac{4\pi }{3}$
• B. $\frac{4\pi }{3\sqrt{3}}$
• C. $\frac{2\sqrt{3}\pi }{3}$
• D. $\frac{2\pi }{3\sqrt{3}}$

Answer 1

Answer: (B)

Given, $arg\left(\frac{z+i}{z-i}\right)=\frac{2\pi }{3}$
Let, $z=x+iy$
$\Rightarrow \frac{x+i\left(y+1\right)}{x+i\left(y-1\right)}=\frac{x^2+\left(y^2-1\right)+2ix}{x^2+{\left(y-1\right)}^2}$
$\Rightarrow ta{n}^{-1}\frac{2x}{x^2+y^2-1}=\frac{2\pi }{3}$
$\Rightarrow x^2+y^2+\frac{2}{\sqrt{3}}x-1=0$
Hence, the given locus is a circle with centre $\left(-\frac{1}{\sqrt{3}},0\right)$ and radius $\frac{2}{\sqrt{3}}$ units

$\Rightarrow$ Length of the arc of the circle is $\frac{2\pi }{3}\times \left(\frac{2}{\sqrt{3}}\right)=\frac{4\pi }{3\sqrt{3}}$ units