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Suppose that $z,w \in \mathbb{C}$ s.t. $z^n = w^n$. I want to prove that $w = z\zeta_n$, where $\zeta_n$ is a n-th root of unity.

What I've tried:

Let $z = r_1e^{i\theta_1}, w = r_2e^{i\theta_2}$.

$z^n = w^n \implies r_1^ne^{i\theta_1 n} = r_2^ne^{i\theta_2 n} \implies r_1 = r_2$ (because $r_1, r_2 > 0$)

and $\theta_1n = \theta_2n + 2k\pi$ for a $k \in \mathbb{Z} $

$ \implies \theta_1 = \theta_2 + \frac{2k\pi}{n}$ for a $k \in \mathbb{Z} \implies z$ is $w$ rotated by an angle $\frac{2k\pi}{n}$ $\implies z = w e^{i\frac{2k\pi}{n}}$,

for a $k \in \mathbb{Z}$, so $z = w\zeta_n$, where $\zeta_n = e^{i \frac{2k\pi}{n}}$ is a n-th root of unity.

Is my proof ok? Thanks!

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    $\begingroup$ "because $r_1,\,r_2>0$" Your stated assumptions don't preclude $z=w=0$, although that edge case works viz. $z=1w,\,1^n=1$. To handle this, begin your proof "without loss of generality $z,\,w$ are not both $0$". $\endgroup$
    – J.G.
    Nov 7, 2023 at 22:01

2 Answers 2

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That's okay.

Another way is $z^n = w^n \implies (z/w)^n = 1$, which means $z/w$ must be equal to a $n$-th root of unity.

Edit: We need to handle $w=0$ separately. But its easy as it implies $z=w=0$ so any number multiplied by either works.

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  • $\begingroup$ math.stackexchange.com/questions/4802537/… I wanted to use this statement here but... I realised that I need a primitive n-th root of unity to work. Any idea for another method? :( $\endgroup$ Nov 7, 2023 at 22:16
  • $\begingroup$ @MathLearner I don't understand how the linked question relates to this. Could you explain a bit more? $\endgroup$
    – whoisit
    Nov 7, 2023 at 22:51
  • $\begingroup$ If your concern is about the implication, I'm using $x=y \implies f(x)=f(y) $ (and not the other way round). Later I'm using the definition of n-th root of unity: Any number $c$ such that $c^n=1$ is an n-th root of unity. $\endgroup$
    – whoisit
    Nov 7, 2023 at 22:53
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Alternative approach:

I am assuming that $~n \in \Bbb{Z^+}.~$

For clarity, I am going to use the variable $~a~$ to represent a (dummy) variable that represents a complex number.

Let $~F~$ denote the set $~\{0,1,2,\cdots,n-1\}.$

For $~k \in F~$ let $~r_k~$ denote $~e^{(i \times 2k\pi/n)}.~$ So, $~r_0, ~r_1, ~\cdots, r_{n-1}~$ represent the $~n~$ distinct roots of the equation $~a^n = 1.~$

Now, consider the equation $~a^n = M ~: ~M \in \Bbb{C}, ~M \neq 0.$

Let $~a_1~$ denote any root of this equation.
Let $~A~$ denote the set $~\{ ~a_1 \times r_k ~: ~k \in F ~\}.$

Then, the set $~A~$ represents the $~n~$ (distinct) roots of the equation $~a^n = M.~$


If $~z^n = 0 = w^n,~$ then you must have that $~z = 0 = w.~$

Otherwise, there exists an $~M \in \Bbb{C},~$ such that $~M \neq 0,~$ and such that $~z^n = M = w^n.~$

This implies (from the first part of this answer) that $~z~$ and $~w~$ are both elements of the set $~A.~$ This implies that there exists $~k \in F~$ such that $~z = w \times r_k.~$

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