$$\because$$ $${z^2} = \overline z \,.\,{2^{1 - |z|}}$$ ...... (1)

$$ \Rightarrow |z{|^2} = |\overline z |\,.\,{2^{1 - |z|}}$$

$$ \Rightarrow |z| = {2^{1 - |z|}}$$,

$$\because$$ $$b \ne 0 \Rightarrow |z| \ne 0$$

$$\therefore$$ $$|z| = 1$$ ...... (2)

$$\because$$ $$z = a + ib$$ then $$\sqrt {{a^2} + {b^2}} = 1$$ ...... (3)

Now again from equation (1), equation (2), equation (3) we get :

$${a^2} - {b^2} + i2ab = (a - ib){2^0}$$

$$\therefore$$ $${a^2} - {b^2} = a$$ and $$2ab = - b$$

$$\therefore$$ $$a = - {1 \over 2}$$ and $$b = \, \pm \,{{\sqrt 3 } \over 2}$$

$$z = - {1 \over 2} + {{\sqrt 3 } \over 2}i$$ or $$z = - {1 \over 2} - {{\sqrt 3 } \over 2}i$$

$${z^n} = {(z + 1)^n} \Rightarrow {\left( {{{z + 1} \over z}} \right)^n} = 1$$

$${\left( {1 + {1 \over z}} \right)^n} = 1$$

$$\left( {{{1 + \sqrt 3 i} \over 2}} \right) = 1$$, then minimum value of n is 6.