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Using the principal complex square root, $\sqrt{i}=e^{i\pi /4}$. Its two square roots differ by a sign, givingEquivalently their arguments are $\pi /8$ and $9\pi /8$ modulo $2\pi$. If instead the right-hand square root is allowed to take both values, the union has four elements $e^{i(\pi /8+k\pi /2)}$, $k=0,1,2,3$, satisfying $z^4=i$. Only the two boxed values square to the principal right-hand side.

Use the principal complex square root, $\sqrt{z}=\sqrt{r}{e}^{i\theta /2}$ with $-\pi <\theta <\pi$. For $F(z)=e^{z-\sqrt{z}}/z$, the local expansion on this holomorphic branch isThe fractional power prevents a single-valued Laurent series in a punctured disc. In particular, zero is a branch point, not an isolated singularity classified as a pole. Merely observing that $zF(z)\to 1$ on the slit plane would not establish a simple pole: $zF(z)$ does not extend as a holomorphic function through a whole disc around zero.

The negative real half-line is a chosen branch cut. Its upper and lower boundary values differ because $\sqrt{-x+i0}=i\sqrt{x}$ and $\sqrt{-x-i0}=-i\sqrt{x}$ for $x>0$. Continuation once around zero changes the sign of the complex square root, and hence changes $F$ in general. Other branch cuts are possible; this one defines the principal holomorphic branch.