$z^3= (z+1)^3 \ \left ( \dfrac {z} {z+1}\right ) =1 = \cos 2 n \pi + i \sin 2 n \pi$ $\left ( \dfrac {z+1}{z} \right )^{-3} = \cos 2n \pi + i \sin 2 n \pi \ \left ( \dfrac {z+1}{z} \right ) = (\cos 2 n \pi + i \sin 2 n \pi)^{-\frac {1}{3}} = \cos \dfrac {2n \pi}{3}- i \sin \dfrac {2n\pi}{3} $ $1+\dfrac {1}{z}=\cos \dfrac {2n\pi}{3} - i \sin \dfrac {2n \pi} {3}\ \dfrac {1} {z}= \cos \dfrac {2 n \pi} -1 - i \sin \dfrac {2 n \pi}{3} $ $z= \dfrac {1}{\cos \frac {2n\pi}{3}-1-i \sin \frac {2n\pi}{3} } = \dfrac {1}{\cos 2x-1-i \sin 2x} \ where, \ x=\dfrac {n \pi}{3},$ $z=\dfrac {1}{-2\sin^2 x-i (2 \sin x \cos x)}= \dfrac {1}{2 \sin x(\sin x+ i \cos x)} $ $=-\dfrac {(\sin x - i \cos x)}{2 \sin x (\sin x+ i \cos x)(\sin x- i \cos x)}= - \dfrac {(\sin x - i \cos x)}{2 \sin x (\sin^2 x - i^2 \cos^2 x)}$ $= -\dfrac {1}{2} \dfrac {\frac {\sin x}{\sin x}- \frac {i\cos x}{\sin x}}{\sin^ x + \cos^2 x}= -\dfrac {1}{2} (1-i \cot x)$ $z=-\dfrac {1}{2}+ \dfrac {i}{2} \cot \left ( \dfrac {n\pi}{3} \right ) \ \therefore, z= - \dfrac {1}{2}+ \dfrac {i}{2} \cot \dfrac {\theta}{2} \ where \ \theta =\dfrac {2n \pi}{3}$