If $\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$, $\;$ then prove that $\;$ $\left|z\right| = 1$
Let $z = x + iy$
Then, $z -1 = \left(x - 1\right) + iy$ $\;$ and $\;$ $z + 1 = \left(x + 1\right) + iy$
Now, $\;$ $arg \left(z - 1\right) = \tan^{-1} \left(\dfrac{y}{x - 1}\right)$ $\;$ and $\;$ $arg \left(z + 1\right) = \tan^{-1} \left(\dfrac{y}{x + 1}\right)$
Given $\;\;$ $arg \left(z -1\right) = \dfrac{\pi}{6}$ $\;$ and $\;$ $arg \left(z + 1\right) = \dfrac{2 \pi}{3}$
$\implies$ $\tan^{-1} \left(\dfrac{y}{x - 1}\right) = \dfrac{\pi}{6}$; $\;$ $\tan^{-1} \left(\dfrac{y}{x + 1}\right) = \dfrac{2\pi}{3}$
$\implies$ $\dfrac{y}{x - 1} = \tan \left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$; $\;$ $\dfrac{y}{x + 1} = \tan \left(\dfrac{2 \pi}{3}\right) = - \sqrt{3}$
i.e. $\;$ $x - \sqrt{3} y = 1$ $\;\;\; \cdots \; (1)$; $\;\;$ $\sqrt{3}x + y = -\sqrt{3}$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously we have,
$4x = -2$ $\implies$ $x = - \dfrac{1}{2}$
Substituting the value of $x$ in equation $(2)$ we have,
$y = - \sqrt{3} + \dfrac{\sqrt{3}}{2} = - \dfrac{\sqrt{3}}{2}$
$\therefore$ $\;$ $z = x + iy = \dfrac{-1}{2} - \dfrac{\sqrt{3}}{2}y$
$\therefore$ $\;$ $\left|z\right| = \sqrt{\left(\dfrac{-1}{2}\right)^2 + \left(\dfrac{-\sqrt{3}}{2}\right)^2}$
i.e. $\;$ $\left|z\right| = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} = 1$