JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Complex Numbers

Taking modulus of both sides of $z^2=\bar{z}\cdot 2^{1-|z|}$: $|z{|}^2=|z|\cdot 2^{1-|z|}$, which gives $|z|=2^{1-|z|}$. Solving this yields $|z|=1$. Substituting back: $z^2=\bar{z}$, so $z^3=z\bar{z}=|z{|}^2=1$. Thus $z$ is a cube root of unity with $b\ne 0$, so $z=\omega$ or $z={\omega }^2$ where $\omega =e^{2\pi i/3}$. We need $z^n=(z+1{)}^n$. Since $1+\omega =-{\omega }^2$, we have ${\omega }^n=(-{\omega }^2{)}^n=(-1{)}^n{\omega }^{2n}$. This requires ${\omega }^{n-2n}=(-1{)}^n$, i.e., ${\omega }^{-n}=(-1{)}^n$. The least $n\in \mathbb{N}$ satisfying this is $n=6$.