Question

Let $z=a+ib,b\ne 0$ be complex numbers satisfying $z^2=\overline{z}.2^{1-|z|}$. Then the least value of $n\in N$, such that $z^n=(z+1{)}^n$, is equal to.

Answer 1

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