Let z = a + ib, b ≠ 0 be complex numbers satisfying z ^ 2 = ¯ z · 2...

Given,

${z}^{2} = \overset{¯}{z} \cdot {2}^{1 - \left|z\right|} ⋯ \left(1\right)$

On taking modulus both side we get,

$⇒ {\left|z\right|}^{2} = \left|z\right| \cdot {2}^{1 - \left|z\right|}$

$⇒ \left|z\right| = {2}^{1 - \left|z\right|}, ∵ b \ne 0 ⇒ \left|z\right| \ne 0$

So comparing both side we get $\left|z\right| = 1 ⋯$\left(2\right)$</p><p>Now putting$z = a + ib$then$\sqrt{{a}^{2} + {b}^{2}} = 1 ⋯ \left(3\right)$</p><p>Now again from equation \left(1\right)$, equation \left(2\right)$, equation \left(3\right)$ we get:

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

Now on comaparing imagenary and real part we get,

$∴ {a}^{2} - {b}^{2} = a$ and $2ab = - b$

Now solving we get, $a = - \dfrac{1}{2}$ and $b = \pm \dfrac{\sqrt{3}}{2}$

So, $z = - \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ or $= - \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$

Now solving ${z}^{n} = {\left(z + 1\right)}^{n} ⇒ {\left(\dfrac{z + 1}{z}\right)}^{n} = 1$

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

$⇒ {\left(\dfrac{1 + \sqrt{3}i}{2}\right)}^{n} = 1$

$⇒ {\left(- {\omega}^{2}\right)}^{n} = 1$, then minimum value of $n$ is $6$