Are there infinite solutions to the complex equation $z^x =27$ when $x$ is irrational?

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Example: $$z^\pi =27$$ with $z \in \mathbb{C}$

$$z=27^{\left(\frac{1}{\pi}\right)}=(r \, e^{i\phi+k2\pi i})^{\left(\frac{1}{\pi}\right)}=r \, e^{\frac{i\phi}{\pi}+k2 i}$$ for $k \in \mathbb{Z}$.

I believe this should mean that the equation has infinite solutions because $$\frac{i\phi}{\pi} + 2 \, k \, i \neq t \pi$$ for $t \in \mathbb{Z}$. Am I correct?