Let z=a+ib, b0, be a complex number satisfying z^2= z2^1-|z|.

Question

Let $z=a+ib,\ b
eq0$, be a complex number satisfying $z^2=\bar z\cdot2^{1-|z|}$. Then the least value of $n\in\mathbb N$ such that $z^n=(z+1)^n$, is equal to \_\_\_\_\_\_\_\_

Answer: 6

Accepted Answer

Taking modulus: $|z|^2=|z|\cdot2^{1-|z|} \implies |z|=2^{1-|z|}$ (since $z\neq0$ as $b\neq0$). This is satisfied by $|z|=1$.

With $|z|=1$: $z^2=\bar z\cdot2^0=\bar z$. Writing $z=a+ib$:
$$(a+ib)^2=a-ib \implies a^2-b^2=a \quad\text{and}\quad 2ab=-b.$$

Since $b\neq0$: $a=-\dfrac12$. Then $\dfrac14-b^2=-\dfrac12 \implies b^2=\dfrac34 \implies b=\pm\dfrac{\sqrt3}2$.

So $z=-\dfrac12+\dfrac{\sqrt3}2i=\omega$ or $z=-\dfrac12-\dfrac{\sqrt3}2i=\omega^2$ (primitive cube roots of unity).

$$z^n=(z+1)^n \implies \left(\frac{z+1}{z}\right)^n=1.$$

With $z=\omega$: $z+1=1+\omega=-\omega^2$, so $\dfrac{z+1}z=\dfrac{-\omega^2}\omega=-\omega$. We need $(-\omega)^n=1$, requiring $n$ even and $3\mid n$ --- the least such $n$ is $6$.

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