$\frac35+i\frac45$ is not a root of unity though its absolute value is $1$.
Suppose I don't have a calculator to calculate out its argument then how do I prove it?
Is there any approach from abstract algebra or can it be done simply using complex numbers?
Any help will be truly appreciated.
$(3+4i)^2 = -7 + 24i = (3+4i)+5(-2+4i)$
With this you can easily prove by induction that for every positive integer $n \ge 1$ there are integers $a_n,b_n$ such that $(3+4i)^n = (3+4i) + 5(a_n+b_n i)$.
In particular, since $4$ is not a multiple of $5$, its imaginary part cannot vanish, so $(3+4i)^n$ can't be a real number when $n \ge 1$.
Here we were lucky that $(3+4i)^n$ mod $5$ is $1$-periodic (that $3+4i = 1 \pmod {2-i}$), in general you can get a cyclic behaviour with larger periods, so it can be a bit longer to write down, but the method works just as well.
Suppose $(a+ib)/c$ is on the unit circle, with $a$ and $b$ coprime.
First, note that if $c$ is even then so are $a$ and $b$ (because $a^2+b^2=c^2$ and squares are congruent to $0$ or $1$ mod $4$), and we supposed they weren't, so $c$ must be odd.
Also, $a$ and $b$ are coprime with $c$ (if $a$ shared a prime factor with $c$, $b$ would also have it).
Now we look at things modulo $(c)$. Let $j \in \Bbb Z/(c)$ such that $a+jb=0$ (this has a unique solution because $b$ is invertible mod $c$).
Then $j^2 = -1 \pmod c$ ($0 = (a+jb)(a-jb) = a^2-j^2b^2$ and since $a^2=-b^2$ and $-b^2$ is invertible, we must have $j^2+1 = 0$), and we can check that the map $\phi : \Bbb Z[i]/(c) \to (\Bbb Z/(c))^2$ defined by $\phi(x+iy) = (x+jy,x-jy)$ is a ring morphism.
If $\phi(x+iy)=0$ then $x+jy=x-jy=0$ from which you get $2x=2y=0$. But $c$ is odd so $2$ is invertible, and so this implies $x=y=0$. This shows that $\phi$ is injective, and since we're dealing with finite rings, it is an isomorphism of rings.
(note that there can be many isomorphisms between those two rings, but this one in particular is the useful one to study $(a+ib)^n$)
Now we have $\phi(a+ib) = (a+jb,a-jb) = (0,a-jb) = (0,2a)$ : Multiplying by $(a+ib)$ is annihilating the first component and multiplying the second by $2a$. And so $\phi((a+ib)^n) = (0^n, (2a)^n)$.
Since $2a$ is invertible mod $c$, the second component is periodic. The first component stays at $0$ from $n \ge 1$, and so the sequence $(\phi((a+ib)^n))_{n \ge 1}$ is periodic. Since $\phi$ is an isomorphism, the sequence $((a+ib)^n \pmod c)_{n \ge 1}$ is also periodic.
Conjugation in $\Bbb Z[i]/(c)$ is translated into the automorphism $(u,v) \mapsto (v,u)$ of $(\Bbb Z/(c))^2$. An element is "real mod $c$" if and only if it is fixed by conjugation. In the $\Bbb Z[i]/(c)$ world this is when the coefficient of $i$ is $0$ (as usual), in the $(\Bbb Z/(c))^2$ world this is when both components are equal.
In particular, since the first component of $\phi((a+ib)^n)$ for $n \ge 1$ is always $0$, it can only be real mod $c$ if both components are $0$. But (unless $c=\pm 1 $ !!) this is impossible because $2a$ is invertible and so its powers can't be zero. So $(a+ib)^n$ is real mod $c$ only when $n=0$ and never again.
Equivalently, this means that the imaginary part of $(a+ib)^n$ is never again a multiple of $c$, and this implies that $(a+ib)^n$ is never real again.
The minimal polynomial of $\frac35+\frac45i$ is quadratic. If it were a primitive $n$th root of unity, the degree should be $\phi(n)$. Hence only few $n$ are possible ($n=3$, $n=4$), but of course this is neither a third nor a fourth root of unity.
If $(3+4i)^n=5^n$ then, in $\mathbb Z[i]$, we have $(2+i)^{2n}=(2+i)^n(2-i)^n$ hence $(2+i)^{n}=(2-i)^n$.
But $2+i$ (and $2-i$, but we don't need it) is a prime element in $\mathbb Z[i]$ (why?), so we must have $2+i\mid 2-i$, impossible. (If $2-i=(2+i)(m+ni)$, then $2=2m-n$ and $-1=m+2n$, therefore $m=\frac35\notin\mathbb Z$.)
Write $e^{i\theta} = \frac{3}{5} + i\frac{4}{5}$. Then $e^{ni\theta} = 1$ only if $\cos(n\theta) = 1$, that is, if $$\cos(n\theta) = T_n(\cos\theta) = T_n\left(\frac{3}{5}\right) = 1$$ where $T_n$ is the $n$-th Chebyshev polynomial.
But the leading term of $T_n$ for $n > 0$ is $2^{n-1}$, thus the minimum denominator of any rational root of $T_n - 1$ must divide $2^{n-1}$. Since 5 does not, $T_n\left(\frac{3}{5}\right) - 1 \neq 0$ for all $n > 0$.
Let $a=\frac{3}{5}+\frac{4}{5}i$. If $a$ were a root of unity, there would exist a positive integer $n$ so that
$$a^n-1=0.$$
Expressions of the form $x^n-1$ can be factored into products of cyclotomic polynomials:
$$x^n-1=\prod_{d|n}\Phi_d(x).$$
There are several properties of cyclotomic polynomials that are relevant here:
The minimal monic polynomial $p(x)\in\mathbb{Q}[x]$ for $a$ is readily calculated to be
$$p(x)=x^2-\frac{6}{5}x+1.$$
In other words, $p(x)$ is the lowest-powered monic polynomial with rational coefficients where $p(a)=0$. Because of the non-integer $-\frac{6}{5}$ coefficient, $p(x)\neq\Phi_d(x)$ for any positive integer $d$ and therefore $a$ is not a root of unity.
We use the formula for $\tan n t$ in terms of $\tan t$ (when $n\in\mathbb N$).
(1) For odd prime $p$, let $ q=(p-1)/2.$ Let $x=\tan \pi/p.$ Then $0=\tan \pi=\tan p(\pi/p) = A(x)/B(x)$ where, for any $z,$ $$A(z)=\sum_{j=0}^q(-1)^j z^{2 j+1}\binom {p}{2 j + 1}$$ $$\text {and }\quad B(z)=\sum_{j=0}^q(-1)^j z^{ j}\binom {p}{2 j}.$$ By Eisenstein's Criterion, the polynomial $A(z)$ is irreducible over $\mathbb Q$, and $\deg(A(z))=p>1$, so any solution of $A(z)=0$ is irrational. Since $A(\tan \pi/p)=A(x)=0,$ we have $$\tan \pi/p\not \in\mathbb Q.$$
(2) Assume $\tan \pi m/n \in\mathbb Q,$ where $m, n\in\mathbb Z^+$, with $\gcd (m,n)=1$, and $n$ is divisible by an odd prime $p.$ Let $r\in\mathbb Z$ such that $m r\equiv 1 \pmod n.$ Then $m r\pi/n=\pi (k+1/n)$ where $k\in Z$, so $\tan m r \pi/n=\pm \tan \pi/n.$
By the formula for $\tan (\pi m/n)r$ in terms of $\tan \pi m/n,$ we have $\tan \pi/n\in\mathbb Q.$ Let $n=p s .$ By the formula for $\tan (\pi/n)s$ in terms of $\tan \pi/s,$ we have $\tan \pi/p\in\mathbb Q.$ This contradicts (1). Therefore $\tan \pi m/n\not \in\mathbb Q$.
(3) We can also show that $\tan m 2^{-m}\pi\not \in\mathbb Q$ when $m,n \in\mathbb N$ with odd $m$ and $n\geq 3$.
(4) Therefore if $z=\cos T+i \sin T$ is an $n$th root of $1$, for some $n\in \mathbb N$, and $z\not \in \{\exp (\pi i j/4) :j\in N\}$ then $\tan T\not \in\mathbb Q.$
In particular, if $\tan T=4/3,$ then $T\ne q\pi$ for any $q\in\mathbb Q,$ so $(3+4 i)/5$ is not an $n$th root of $1.$
In $\mathbb Z_5[i]$ : $(3+4i)(3+4i) = 3+4i$
$\forall n \ge 1$ : $(3+4i)^n \notin \Re$
$(3+4i)/5$ is not a root of unity.
