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The answer to your question follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable (in sense of your question) iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f_x(\sin x)$, where function $f_x$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.

Let's find all $n$ such that $\forall a,b \in R \Rightarrow z^n =a+ib$ is positive solvable. The answer follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable $\forall a,b\in R$ iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f_x(\sin x)$, where function $f_x$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.

The answer to your question follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable (in sense of your question) iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f(\sin x)$, where function $f$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.

The answer to your question follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable (in sense of your question) iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f_x(\sin x)$, where function $f_x$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.

The answer to your question follows directly from the paper mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable (in sense of your question) iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f(\sin x)$, where function $f$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.

The answer to your question follows directly from this paper by Brian Conrad, mentioned above. Some information can also be found here: Weisstein, Eric W. "Trigonometry Angles."

First let's observe that $z^n =a+ib$ is positive solvable (in sense of your question) iff $\forall x \in R$: $\sin\displaystyle\Big(\frac{x}{n}\Big)=f(\sin x)$, where function $f$ has an algebraic expression over the rationals involving rational powers. Let's call the set of all such positive integers $\Omega$. Let's prove that $\Omega=\{2^k\}$. We know (from the papers mentioned above) that $\sin\displaystyle\Big(\frac{\pi}{n}\Big)$ can be expressed in radicals iff $\varphi(n)=2^m$ where $\varphi$ is Euler's totient function. Then we observe that the only prime $p$ such that $\varphi(p^2)=p(p-1)=2^k$ is $p=2$. Assume that some other prime $p\in\Omega$. From definition of $\Omega$ follows $p^2\in\Omega$, than we have a contradiction taking $x=\pi$. Now let's assume that some $n=p_1p_2...p_n \in \Omega$ and prime $p_1\ne2$. Then applying $\sin(x+y)$ formula to $\sin\displaystyle\Big(p_2p_3...p_n\frac{x}{n}\Big)= \sin\displaystyle\Big(\frac{x}{p_1}\Big)$ we obtain $p_1\in\Omega$ which again leads to contradiction. So we get $\Omega\subset\{2^k\}$. Inverse embedding is obvious.