TITLE: A case where $z^z = 0$ where $z$ is complex number
QUESTION [5 upvotes]: Is there any case where $z^z = 0$ where $z$ is complex number? The case excludes the case where $z=0$.

REPLY [16 votes]: No. If $z\ne 0$ we have by definition $z^z=e^{z\log z}$, with some arbitrary choice of $\log z$ made (which is only unique up to a multiple of $2\pi i$). But $e^w$ is nonzero for all complex numbers $w$.