No. 1!

$Let\quad z=a+\iota b\\ \left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } -(1)\\ \left| z+1 \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 }+2a+1 } -(2)\\ Solving\quad (1)=1\quad \& \quad (2)=1\\ we\quad get\quad that\quad a=-\frac { 1 }{ 2 } \quad \& \quad b=\frac { \sqrt { 3 } }{ 2 } \\ so\quad z=-\frac { 1 }{ 2 } +\iota \frac { \sqrt { 3 } }{ 2 } \\ i.e\quad z=\omega \quad or\quad z={ e }^{ \iota 2\pi /3 }\quad and\quad z=\omega +1or\quad z={ e }^{ \iota \pi /3 }\\ Using\quad the\quad properties\quad of\quad \omega \quad or\quad finding\quad the\\ value\quad of\quad n\quad such\quad that\quad \iota \frac { 2\pi }{ 3 } =\iota \frac { \pi }{ 3 } \\ we\quad get\quad \boxed{n=6}$

Clearly one of the solutions for z is ω, others being it's exponents . And the second equation only suggests that difference between the arguments of numbers on either side is 360°. 60*n = 360 implies, n = 6

The solution is w and w^2. (w is omega). As 1+w=-w^2, So w^n=(-w^2)^n. So minimum value of n is 6.