Let $n_{0},n_{1} \in \mathbb{N} $ such that $1< n_0,n_1$ and $gcd(n_0,n_1)=1$. Let $w_{0},w_{1} \in \mathbb{C}$.
Conjecture:
if there is a number $z \in \mathbb{C}$ which satisfies $$\begin{cases} z^{n_{0}}=w_{0} \\ z^{n_{1}}=w_{1} \\ \end{cases}$$ then $z$ is unique; or in other words, if there is a solution to the above system then it is the only solution.
Is the conjecture true?
My work so-far:
Suppose $z \in \mathbb{C}$ satisfies the above system. If WLOG $\Im(w_{0})=0$ then $z=w_{0}^{\frac{1}{n_{0}}}$ thus $\Im(w_{1})=0$ and $z$ is unique. otherwise, if $\Im(w_{0}),\Im(w_{1})\neq 0$, then assume WLOG (justified by the former case) that $|w_{0}|=m^{\frac{1}{n_{0}}}$ and $|w_{1}|=m^{\frac{1}{n_{1}}}$ for some $ m \in \left( 0, \infty \right)$, then use de Moivre's theorem to obtain: $$\begin{cases} z=m\exp\left(i\left(\theta_{0}+2\pi k_{0} \right)/n_{0} \right)&\text{for $k_{0} \in \{0,1,...,n_{0}-1 \}$ and $\theta_{0} \in \left[0,2 \pi \right)$} \\ z=m\exp\left(i\left(\theta_{1}+2\pi k_{1} \right)/n_{1} \right)&\text{for $k_{1} \in \{0,1,...,n_{1}-1 \}$ and $\theta_{1} \in \left[0,2 \pi \right)$} \end{cases}$$ Denote $\alpha_{0} \equiv \theta_{0}/2\pi$ and $\alpha_{1} \equiv \theta_{1}/2\pi$ for $\alpha_{0},\alpha_{1} \in \left[0,1\right)$ to obtain: $$\begin{cases} z=m\exp\left(i2\pi\left(\left(\alpha_{0}+ k_{0} \right)/n_{0}\right) \right)&\text{for $k_{0} \in \{0,1,...,n_{0}-1 \}$ and $\alpha_{0} \in \left[0,1 \right)$} \\ z=m\exp\left(i2\pi\left(\left(\alpha_{1}+ k_{1} \right)/n_{1}\right) \right)&\text{for $k_{1} \in \{0,1,...,n_{1}-1 \}$ and $\alpha_{1} \in \left[0,1 \right)$} \end{cases}$$ hence $$\frac{\alpha_{0}+ k_{0}}{n_{0}}=\frac{\alpha_{1}+ k_{1}}{n_{1}} $$ Rearrange the above to obtain: $$n_{1}k_{0}-n_{0}k_{1}=n_{0}\alpha_{1}-n_{1}\alpha_{0} $$ thus denote $j \equiv n_{0}\alpha_{1}-n_{1}\alpha_{0}$ to obtain that $ j \in \mathbb{Z} $. The diophantine equation $n_{1}k_{0}-n_{0}k_{1}=j$ has a solution because $gcd(n_0,n_1)=1$. If $\left( k_{0}, k_{1} \right)$ is a solution of the diophantine equation then every other solution is of the form $\left( k_{0}+an_{0}, k_{1}-an_{1} \right)$ for some $a \in \mathbb{Z}$. Denote $k_{0}' \equiv k_{0}+an_{0}$ and $k_{1}' \equiv k_{1}-an_{1}$; if $a=0$ then $\left( k_{0}, k_{1} \right)=\left( k_{0}', k_{1}' \right)$, otherwise if $a \neq 0$ then $\left( k_{0}', k_{1}' \right) \notin \{0,1,...,n_{0}-1 \} \times \{0,1,...,n_{1}-1 \} $ - thus the solution $\left( k_{0}, k_{1} \right)$ is unique hence $z$ is unique.
Is the above correct? Is there a more linear-algebraic argument than the above? Is there a more general solution?
EDIT: as the comment by man o shadow says - the first line of my solution is incorrect. If $ z \in \left[ 0, \infty \right)$ then $z=w_{0}^{1/n_{0}}$ and thus $z$ is unique; otherwise if $ z \in \mathbb{C} \setminus \mathbb{R}_{\ge 0}$ then assume WLOG (justified by the former case) that $|w_{0}|=m^{\frac{1}{n_{0}}}$ and $|w_{1}|=m^{\frac{1}{n_{1}}}$ for some $ m \in \left( 0, \infty \right)$...
The first line in your argument is incorrect. Even is $w_0$ is real the quantity $w_0^{\frac{1}{n_0}}$ is, in general, not well defined as the polynomial $z^{n_0}-w_0$ will have $n_0$ roots counting multiplicity. Just think of $z^n-1$ which has $n$ distinct roots, even if 1 is real.
As for an alternative approach. Since $n_0$ and $n_1$ are relatively prime, there exist integers $a$ and $b$ such that $an_0+bn_1=1$ so we have that $z=z^{an_0+bn_0}=(z^{n_0})^a(z^{n_1})^b=w_0^aw_1^b$ and this tells you z in terms of $w_0$ and $w_1$.