Question

If $z$ is a complex number such that $\arg \left(\frac{z-1}{z+1}\right)=\frac{\pi }{3}$, then find the value of $\sqrt{3}\left|z-\frac{1}{\sqrt{3}}i\right|$.

Answer 1

Answer: 2

If $z=x+iy$, then
$\mathrm{Arg}(z-1)={\tan }^{-1}\left(\frac{y}{x-1}\right)$
$\mathrm{Arg}(z+1)={\tan }^{-1}\left(\frac{y}{x+1}\right)$
$\mathrm{Arg}(z-1)-\mathrm{Arg}(z+1)={\tan }^{-1}\left(\frac{y}{x-1}\right)-{\tan }^{-1}\left(\frac{y}{x+1}\right)$
$\frac{\pi }{3}={\tan }^{-1}\left(\frac{2y}{1+x^2+y^2}\right)\Rightarrow x^2+y^2-\frac{2}{\sqrt{3}}y=1\Rightarrow x^2+{\left(y-\frac{1}{\sqrt{3}}\right)}^2=\frac{4}{3}$
$\Rightarrow {\left|z-\frac{i}{\sqrt{3}}\right|}^2=\frac{4}{3}\Rightarrow \sqrt{3}\left|z-\frac{i}{\sqrt{3}}\right|=2$