If $\lambda = e^{i\mu}$ and $\mu \simeq 0$, can we have $\lambda^{k} = 1$ for some $k <5$?

Question

Suppose I have $\lambda = e^{i\mu}$, where $\mu = \mu(\epsilon)$ with $\epsilon$ a small parameter. Suppose $\mu$ can be expanded in series in $\epsilon$ as $\mu = \mu_{0} + \epsilon \mu_{1} + O(\epsilon)^{2}$, with $\mu_{0}=0$. In other words, for small $\epsilon$, $\mu$ tends to $0$.

Is it possible for $\lambda^3 =1 $ or $\lambda^{4} = 1$ when taking the exact value of $\mu$ for some small $\epsilon$? More generally, can we have $\lambda^{k} = 1$ for $k<5$ ?

Answer 1

$$1=\lambda^k=e^{ik\mu}\implies\mu=\frac{2n\pi}k$$

So for small enough $\epsilon>0$, since $\mu\to0$ but $\mu\ne0$, then $\lambda^k\ne1$.