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Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.

by contradiction $\exists n$ s.t. $(\frac{3}{5}+\frac{4}{5}i)^n=1$

$(3+4i)^n=5^n$

when $n=2$, $(3+4i)^2=3+4i\pmod5$

stuck how prove that $(3+4i)^n$=$3+4i\pmod5$ use induction? if yes how in this case?

for second part of the question again by contradiction.

$\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$

$\phi=\arctan(\frac{4}{3})=\frac{\pi m}{n}$ how continue from here?

miracle173
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unit 1991
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  • Tell us where you came across this question. There are multiple ways to answer and which way is most helpful to you depends on what you are expected to know (but in any case, Hint: the two questions are related). – Sean Eberhard Mar 08 '22 at 09:25
  • @SeanEberhard I came across to this question in group theory class. I know basic things like order of a group,subgroup. – unit 1991 Mar 08 '22 at 09:27
  • @SeanEberhard They are asked in one question so I assumed they are related:) – unit 1991 Mar 08 '22 at 09:27

2 Answers2

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As you observed, $(3+4i)^2 = -7 + 24i \equiv 3 + 4i \pmod 5$. Therefore $$(3+4i)^n =(3+4i)^{n-2} (3+4i)^2 \equiv (3+4i)^{n-2} (3+4i) = (3+4i)^{n-1} \pmod 5$$ for all $n > 1$, so $(3+4i)^n \equiv 3+4i \pmod 5$ by induction. Hence $(3+4i)^n \neq 5^n$ for $n > 0$.

If you know some algebraic number theory, you can say the following. We can write $z = (3+4i)/5 = (2+i)/(2-i)$, and $2+i$ and $2-i$ are primes in $\mathbf{Z}[i]$, which is a UFD, so $z^n = 1$ is impossible by unique factorization.

I will leave the second part to you. Hint: write $(3+4i)/5$ in polar form.

Sean Eberhard
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For the second part of the question: O.P. assumed that $\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$ is rational. Then, $\cos(\frac{m}{n}\pi)=\frac{3}{5}$.

Let $U_{n-1}(x)$ be the Chebyshev polynomial of second kind of order $n-1$. Then $U_{n-1}(\frac{3}{5})=\frac{\sin m\pi)}{\sin(\frac{m}{n}\pi)}=0.$ We will show that this is impossible giving us a contradiction.

From https://en.wikipedia.org/wiki/Chebyshev_polynomials, I got $$U_{n}(x)=\frac{(x+\sqrt{x^2-1})^n-(x-\sqrt{x^2-1})^n}{2\sqrt{x^2-1}}.$$ I am not sure if this is for all $x$. I hope so. Then, when we put $x=\frac{3}{5}$ we get $$(3+4i)^{2n}=5^{2n}$$ which is not possible due to the answer of S. Eberhard.

Bob Dobbs
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