From the fact that $\frac{1}{5}(3+4i)$ has infinite order in $(\mathbb{C},\cdot)$, I'm supposed to infer that $\frac{1}{\pi}\arctan{\frac{4}{3}}$ is irrational. Irrationality of $\arctan\frac{4}{3}$ follows immediately but I can't see why the irrationality of the product does. Any hints would appreciated!
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1If $\frac{1}{\pi}\arctan{\frac{4}{3}}$ is rational then $\arctan{\frac{4}{3}}=\pi\frac mn$ and at some integral power it is an even multiple of $\pi$. – user26857 Oct 12 '17 at 15:10
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see this post that contains proof that arctan of any rational is not a rational multiple of $\pi$ https://math.stackexchange.com/questions/79861/arctan2-a-rational-multiple-of-pi/79867#79867 – Vasili Oct 12 '17 at 15:21
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Let $\alpha$ be your complex number, then $$\alpha^n = \frac{a_n}{5^{n+1}} + \frac{b_n}{5^{n+1}}i$$
and since we have $$\alpha^n = \frac{1}{5}(3+4i)\left(\frac{a_{n-1}}{5^n} + i\frac{b_{n-1}}{5^n}\right) = \frac{1}{5^{n+1}}\left(3a_{n-1} - 4b_{n-1} + i(a_{n-1} + 3b_{n-1})\right)$$ it is clear that $a_n =3a_{n-1} - 4b_{n-1}$ and $b_n = a_{n-1} + 3b_{n-1}$, with $a_1 = 15$ and $b_1 = 20$.
Examining everything $\pmod{7}$ we find $(a_1, b_1) = (1, 6), (a_2, b_2) = (0, 5), (a_3,b_3) = (6,1)$ and $(a_4,b_4)=(0,2), (a_5, b_5) = (1,6)$ so the cycle continues with no $b_n = 0$.
Hence $\alpha^n$ is never real, but if $\arctan(4/3) = \frac{p}{q}\pi$ we would have $\exp\left(\frac{p}{q}\pi i \cdot{2q}\right) = 1 = \alpha^{2q}$.
Zain Patel
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