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## New answers tagged tetration

2

No. If $|z|<1$ the limit is $0$, if $|z|>1$ the limit is $\infty$ or does not exist since $|z|^n\to\infty$. If $|z|=1$ then $z=e^{i\theta}$ for $\theta\in[0,2\pi)$ and $z^n=e^{in\theta}$. Unless $\theta=0$, i.e. $z=1$, the sequence will rotate the point on the unit circle by $\theta$ counterclockwise at each step, and there is no limit.

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