Prove or disprove $\frac{1}{5}(3-4i)$ is a root of unity.
Here is the definition of root of unity:
An nth root of unity, where $n$ is a positive integer $(i.e. n = 1, 2, 3, …)$, is a number $z$ satisfying the equation $z^n=1$.
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2We can also say that $2\pi$ is not an integral multiple of $\arctan(\frac{4}{3})$. – jonsno Jun 17 '17 at 06:36
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There are many ways to do this; roots of unity are algebraic integers but this number $\alpha$ isn't. Or you could show the denominator of $\alpha^n$ is always $5^n$ etc.
Angina Seng
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It is not a root of unity. Since $(3+i)(3+i)=8+6i\equiv 3+i \pmod{5}$ it follows that for any positive integer $n$, $$(3-4i)^n\equiv (3+i)^n\equiv 3+i \pmod{5}$$ which contradicts the fact that $(3-4i)^m=5^m$ for some positive integer $m$.
Robert Z
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Why do we need to say "since $(3+i)(3+i)=...$? I think we can directly conclude that $(3-4i)^n$ is congruent to $(3+i)^n$ because $-4=1$ modulo $5$? – zero field Jun 17 '17 at 07:11
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@zero field The "since ..." is needed to show that $(3+i)^n\equiv 3+i$ – Robert Z Jun 17 '17 at 07:20
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Consider the exponential form.
We have $\displaystyle q=e^{i\tan^{-1}\left(-\frac{4}{3}\right)}$
All roots of unity are of the form $z=e^{i\left(\frac{2π}{n}\right)}$.
This is because they are all solutions of $z^n=1$.
Notice that $\tan^{-1}(-\frac{4}{3})$ is not in this form, and thus $q$ is not a root of unity.
Saketh Malyala
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To be more specific, all roots of unity are of the form $z=e^{i\left(\frac{2kπ}{n}\right)}$ where $1\leq k<n$ – Serenity Jun 17 '17 at 06:52