Let $\mathrm{z}=a+i b, b \neq 0$ be complex numbers satisfying $z^{2}=\bar{z} \cdot 2^{1-z}$. Then the least value of $n \in N$, such that $z^{n}=(z+1)^{n}$, is equal to __________.

<p>$\because$ ${z^2} = \overline z \,.\,{2^{1 - |z|}}$ ...... (1)</p> <p>$\Rightarrow |z{|^2} = |\overline z |\,.\,{2^{1 - |z|}}$</p> <p>$\Rightarrow |z| = {2^{1 - |z|}}$,</p> <p>$\because$ $b \ne 0 \Rightarrow |z| \ne 0$</p> <p>$\therefore$ $|z| = 1$ ...... (2)</p> <p>$\because$ $z = a + ib$ then $\sqrt {{a^2} + {b^2}} = 1$ ...... (3)</p> <p>Now again from equation (1), equation (2), equation (3) we get :</p> <p>${a^2} - {b^2} + i2ab = (a - ib){2^0}$</p> <p>$\therefore$ ${a^2} - {b^2} = a$ and $2ab = - b$</p> <p>$\therefore$ $a = - {1 \over 2}$ and $b = \, \pm \,{{\sqrt 3 } \over 2}$</p> <p>$z = - {1 \over 2} + {{\sqrt 3 } \over 2}i$ or $z = - {1 \over 2} - {{\sqrt 3 } \over 2}i$</p> <p>${z^n} = {(z + 1)^n} \Rightarrow {\left( {{{z + 1} \over z}} \right)^n} = 1$</p> <p>${\left( {1 + {1 \over z}} \right)^n} = 1$</p> <p>$\left( {{{1 + \sqrt 3 i} \over 2}} \right) = 1$, then minimum value of n is 6.</p>