$Let \ z=x+iy$ $z+1 = (x+1) + iy; z-1 = (x-1)+iy$ $arg \ (z+1) = \dfrac {\pi}{6}$ $\tan^{-1} \left ( \dfrac {y}{x+1}\right ) = \dfrac {\pi}{6}$ $\dfrac {y}{x+1}= \tan \dfrac {\pi}{6}= \dfrac {1}{\sqrt{3}}$ $y=\dfrac {1}{\sqrt{3}} (x+1) \ \cdots \ \cdots (A)$ $arg (z-1) = \dfrac {2\pi}{3}$ $\tan^{-1} \left ( \dfrac {y}{x-1} \right ) = \dfrac {2\pi}{3}$ $\dfrac {y}{x-1}= \tan \dfrac {2\pi}{3}= - \sqrt{3} $ $y =-\sqrt{3} (x-1) \ \cdots \ \cdots (B)$ from (A) and (B), $\dfrac {1}{\sqrt{3}} (x+1) = -\sqrt{3}(x-1)$ $x+1 = -3 (x-1)$ $x+3x=3-1 $ $x=\dfrac {1}{2}$ $y=\dfrac {1}{\sqrt{3}} (x+1) = \dfrac {1}{\sqrt{3}} \left ( \dfrac {1}{2} +1 \right ) = \dfrac {3}{2\sqrt{3}} $ $y=\dfrac {\sqrt{3}}{2}$ $\therefore z=\dfrac {1}{2}+ \dfrac {i\sqrt{3}}{2}$