If $\operatorname{Im}(z) \in\Bbb Q$ for $|z|=1$ then what about real and imaginary parts of $z^{2^n}$

Question

Given in the question that let $z \in \mathbb{C}$ with $|z|=1$ and let imaginary part of $z$ be rational. Then which of the following is necessarily true?

(A) For each $n \ge1$ the imaginary part of $z^{2^n}$ is rational.

(B) For each $n \ge1$ the real part of $z^{2^n}$ is rational.

(C) There exists $n_0$ such that for each $n \ge n_0$ the real and imaginary part of $z^{2^n}$ is rational.

(D) None of the above

My approach is to consider $z=\cos\theta +i\sin\theta$ and $z^{2^n}=\cos {2^n}\theta +i\sin{2^n}\theta$. Neither the real part and nor the imaginary part can be rational for all $n$. So answer should be (D). Please confirm.

Answer 1

If $a,b\in\mathbb{R}$ are such that $|a+bi|=1$, then $a^2+b^2=1$. So, $a=\pm\sqrt{1-b^2}$, and therefore, if $b\in\mathbb{Q}$ and $k$ is an even natural number, $a^k$ is a rational number. On the other hand, $$ \operatorname{Re}\left((a+bi)^n\right)=a^n-a^{n-2}b^2+a^{n-4}b^4-\cdots, $$ and therefore, assuming that $b\in\mathbb{Q}$ and that $n$ is even, $\operatorname{Re}\left((a+bi)^n\right)\in\mathbb{Q}$.

On the other hand, if $z=\frac{\sqrt3}2+\frac i2$, then $\operatorname{Im}\left(z^{2^n}\right)=\pm\frac{\sqrt3}2$, for every natural number $n$.

Answer 2

For option (b): for any complex number $a+bi$ with $|a+bi| = 1$, i.e. $a^2+b^2 = 1$, squaring it gives

$$\begin{align*} (a+bi)^2 &= a^2 - b^2 +2abi\\ &= 1-2b^2 + 2abi \tag 1\\ &= 2a^2-1 + 2abi \tag 2 \end{align*}$$

Using the last two forms, if either the real or the imaginary part of $a+bi$ is rational, then the real part of $(a+bi)^2$ is rational.

By induction, since the imaginary part of $z$ is rational, then the real part of any $\overbrace{\left(\left(\left(z^2\right)^2\right)^\cdots\right)^2}^{n \text{ squaring}} = z^{2^n}$ is rational for $n\ge 1$.

(The real parts in $(1)$ and $(2)$ are also known as

$$\cos 2\theta = 1-2\sin^2\theta = 2\cos^2\theta-1$$

)