Let, $Z=x+iy\ \therefore z+1 = (x+1) + iy \ \text{ and } z-1 = (x-1)+iy $

We are given arg $(z+1) = \dfrac {\pi}{6} \ \therefore \tan^{-1} \dfrac {y}{x+1} = \dfrac {\pi}{6 }$

$\therefore \dfrac {y}{x+1} = \tan \dfrac {\pi}{6} = \dfrac {1}{\sqrt{3}}$

$ \therefore \sqrt{3}y=x+1 \ \cdots \ \cdots (1)$

$\text{Also arg }(z-1)=\dfrac {2\pi}{3}\ \therefore \ \tan^{-1} \dfrac {y}{x-1}=\dfrac {2\pi}{3}$

$\therefore \dfrac{y}{x-1}=\tan \ \dfrac {2\pi}{3}= -\sqrt{3} $

$\therefore -y=\sqrt{3}x-\sqrt{3}$

$-\sqrt{3}y=3x-3\cdots \ \cdots (2)$

Adding (1) and (2) we get

$2x=1, \text{hence} \ x=\dfrac {1}{2}$

Substituting values of x in (1)

$y= \dfrac {\sqrt{3}}{2}$

$\therefore z=x+iy=\dfrac {1}{2}+i \dfrac {\sqrt{3}}{2}$