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If $z$ is a non-zero complex number and $m,n\in\mathbb{Z^+}$ and $gcd(m,n)=1$,then $(z^{\frac{1}{n}})^m=(z^m)^{\frac{1}{n}}$.

The proof of this theorem is given as follows:

Let $z=r(cos\theta+isin\theta)$, where $-\pi<\theta\leq\pi$. Now, $(z^{\frac{1}{n}})^m=r^{\frac{m}{n}}(cos\frac{2k\pi+\theta}{n}+isin\frac{2k\pi+\theta}{n})^m=r^{\frac{m}{n}}(cos\frac{\theta m}{n}+isin\frac{\theta m}{n})(cos\frac{2\pi mk}{n}+isin\frac{2\pi mk}{n})=r^{\frac{m}{n}}(cos\frac{\theta m}{n}+isin\frac{\theta m}{n})\omega^{km}$, where $\omega=cos\frac{2\pi}{n}+isin\frac{2\pi}{n}$ and $k=0,1,2,...,n-1$. Now, $(z^m)^{\frac{1}{n}}=[r^m(cosm\theta+isinm\theta)]^\frac{1}{n}=r^\frac{m}{n}[cos\frac{2k\pi+m\theta}{n}+isin\frac{2k\pi +m\theta}{n}]=r^\frac{m}{n}(cos\frac{m\theta}{n}+isin\frac{m\theta}{n})\omega^k$, where $\omega=cos\frac{2\pi}{n}+isin\frac{2\pi}{n}$ and $k=0,1,2,...,n-1$.Thus, $$(z^m)^{\frac{1}{n}}=r^\frac{m}{n}(cos\frac{m\theta}{n}+isin\frac{m\theta}{n})\omega^k$$ and $$(z^{\frac{1}{n}})^m=r^\frac{m}{n}(cos\frac{m\theta}{n}+isin\frac{m\theta}{n})\omega^{mk}$$. Now, if $S={0,m,2m,3m,..,(n-1)m}$. If $a_i\in S $ and if $a_i$ is divided by $n$, they all leave remainders $0,1,2,...,n-1$.Thus, $a_i=km=nq+r$, where $k=0,1,2,...,n-1$ and $r=0,1,2,...,n-1$ . Hence, $\omega ^{mk}=\omega^{qn}\omega^r=\omega^{qn}\omega^r=(\omega^n)^q.\omega^r$. Now, $\omega^n=1$, $\omega ^{mk}=\omega^r=\omega^k$. Thus, $(z^{\frac{1}{n}})^m=(z^m)^{\frac{1}{n}}$ when $gcd(m,n)=1$.

Now, is the above proof valid ? If so , then why can't we replicate the above proof as it is and claim that $(z^{\frac{1}{n}})^m=(z^m)^{\frac{1}{n}}$ for all cases even when $gcd(m,n)\neq 1$. I am not quite getting it...

Arthur
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    $\ z^\frac{1}{n}\ $ is not uniquely defined, but if $\ z=1\ $, say, then no matter which $\ n^\text{th}\ $ root $\ e^\frac{2\pi ij}{n}\ $ of $\ 1\ $ you choose $\ z^\frac{1}{n}\ $ to be, you will have $\ \left(z^\frac{1}{n}\right)^n=1\ $, whereas $\ \big(z^n\big)^\frac{1}{n}=1^\frac{1}{n}\ $ is not necessarily $1$. It could have any one of the values $\ e^\frac{2\pi ij}{n}\ $ for $\ j=1,2,\dots, 2n-1\ $. – lonza leggiera Oct 18 '22 at 08:05
  • @Ionza leggiera What I was talking about is that how $gcd(m,n)=1$ is relevant in the proof? I mean the fact is not used anywhere in the proof, right? So, can't we replicate the same proof when $gcd(m,n)\neq 1$ ?...Doesn't the same thing holds when $gcd(m,n)\neq 1$ ?... – Arthur Oct 21 '22 at 01:54
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    It's used in the conclusion "If $\ a_i\in S\ $ and if $\ a_i\ $ is divided by $\ n\ $ they all leave remainders $\ 0,1,2,\dots,n-1\ $." This is not true unless $\ \gcd(m,n)=1\ $. In general, the remainders will be $\ 0,d,2d,\dots,n-d\ $, where $\ d=\gcd(m,n)\ $. Nevertheless, the proof is problematic, in my opinion, mainly because of the ambiguity in the statement of what it's trying to prove. – lonza leggiera Oct 21 '22 at 02:57
  • @Ionza leggiera Thank you so much! However, according to Euclid's division algorithm , when $a,b\in \mathbb{Z}$ and if $a$ is divided by $b$ then we can say $a=bq+r$, where $q,r\in \mathbb{Z}$ and $r$ is the remainder and $0\leq r< n$. Isn't it valid for all caeses?...I am not quite getting this... – Arthur Oct 22 '22 at 07:34
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    Yes. But if $\ a\ $ and $\ b\ $ both divisible by $\ d\ $, then $\ r\ $ must also be divisible by by $\ d\ $. It can't therefore take on all the values $\ 0,1,2,\dots,b-1\ $ from $\ 0\ $ to $\ b-1\ $, but only those divisible by $\ d\ $—namely $\ 0,$$,d,$$,2d,$$,\dots,$$,b-d\ $. – lonza leggiera Oct 22 '22 at 08:05
  • @Ionza leggiera Thank you so much! I do get it now...Thank you!... – Arthur Oct 22 '22 at 14:58

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