# If $z^n = w^n$, then z is w multiplied by a root of unity

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!

• "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 $0$".
– J.G.
Nov 7, 2023 at 22:01

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.

• 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? :( Nov 7, 2023 at 22:16
• @MathLearner I don't understand how the linked question relates to this. Could you explain a bit more? Nov 7, 2023 at 22:51
• 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. Nov 7, 2023 at 22:53

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 $$~z = w \times r_k.~$$