Complex World (Correct answer updated)

The equation simplifies to $3z^2+3z+1=0$ . The sum of the roots is $\boxed{-1}$ by Viete.

${ \left( \frac { z }{ z+1 } \right) }^{ 3 }=1=\cos { 0+i\sin { 0=\cos { 2n\pi +i\sin { 2n\pi } } } }$

$\therefore \quad \frac { z }{ z+1 } ={ (\cos { 2n\pi +i\sin { 2n\pi } } ) }^{ \frac { 1 }{ 3 } }=\cos { \frac { 2n\pi }{ 3 } +i\sin { \frac { 2n\pi }{ 3 } } ;\quad (n=0,1,2) }$

$\therefore \quad \frac { z }{ z+1 } =\cos { \theta } +i\sin { \theta } ;\quad \theta =\frac { 2n\pi }{ 3 } ,\quad \quad (n=0,1,2)$

$\therefore \quad \frac { z }{ z+1-z } =\frac { \cos { \theta +i\sin { \theta } } }{ 1-\cos { \theta -i\sin { \theta } } }$

$\therefore \quad z=\frac { \cos { \theta +i\sin { \theta } } }{ 2\sin ^{ 2 }{ \frac { \theta }{ 2 } -2i\sin { \frac { \theta }{ 2 } .\cos { \frac { \theta }{ 2 } } } } }$

$=\frac { \cos { \theta +i\sin { \theta } } }{ 2\sin { { (\theta }/{ 2) } } \left[ \sin { { (\theta }/{ 2)-i\cos { ({ \theta }/{ 2) } } } } \right] } .\frac { \left[ \sin { { (\theta }/{ 2)+i\cos { ({ \theta }/{ 2) } } } } \right] }{ \left[ \sin { { (\theta }/{ 2)+i\cos { ({ \theta }/{ 2) } } } } \right] }$

$=\frac { (\cos { \theta +i\sin { \theta } )\left[ \sin { { (\theta }/{ 2)+i\cos { ({ \theta }/{ 2) } } } } \right] } }{ 2\sin { { (\theta }/{ 2) } } \left[ \sin ^{ 2 }{ \left( { \theta }/{ 2 } \right) +\cos ^{ 2 }{ \left( { \theta }/{ 2 } \right) } } \right] }$

Simplifying further and grouping the real and imaginary parts;

$=\frac { -\left[ \sin { \theta \cos { ({ \theta }/{ 2)-\cos { \theta \sin { \left( { \theta }/{ 2 } \right) } } } } } \right] +i\left[ \cos { \theta \cos { \left( { \theta }/{ 2 } \right) +\sin { \theta \sin { \left( { \theta }/{ 2 } \right) } } } } \right] }{ 2\sin { ({ \theta }/{ 2) } } }$

$\therefore \quad z=\frac { -\sin { ({ \theta }/{ 2)+i\cos { \left( { \theta }/{ 2 } \right) } } } }{ 2\sin { \left( { \theta }/{ 2 } \right) } } =-\frac { 1 }{ 2 } +\frac { i }{ 2 } \cot { \frac { \theta }{ 2 } }$ ; where $\theta =\frac { 2n\pi }{ 3 }$ .

Here $n=0,1,2$ ; but when $n=0$ , imaginary part is not defined.

$\therefore \quad n\neq 0$

$\therefore$ there are only two roots corresponding to $n=1,2$ ,

$\therefore$ sum of the real parts of two roots = $-\frac { 1 }{ 2 } -\frac { 1 }{ 2 } =\boxed{-1}$ .