QJEEM22Mae39f: JEE Main 2022 Mathematics - Class 11 JEEM22Mae39f Solution: Let be complex numbers satisfying . Then the least value of ,…

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