Let f(z)=z^(1/2) . Find f^*(-e) where f^*(z) is a…
Let $f(z)=z^{1/2}$. Find $f^*(-e)$ where $f^*(z)$ is a single-valued branch with $f^*(1)=1$, extended continuously as $z$ moves through: $1^\circ$ the upper half-plane; $2^\circ$ the lower half-plane.
Explanation of Method
Write $z=re^{i\theta}$ and $f(z)=\sqrt re^{i\theta/2}$. The branch point is $z=0$; moving from $z=1$ ($\theta=0$) to $z=-e$ through the upper half-plane versus the lower half-plane forces $\theta$ to reach $+\pi$ or $-\pi$ respectively at $z=-e$, giving two different values.
Step 1 · Setup
At $z=1$: $r=1,\theta=0$, so $f^*(1)=1^{1/2}e^{i\cdot0}=1$ ✓, fixing the branch's normalization.
Part $1^\circ$: path through the upper half-plane
Moving along an arc through the upper half-plane from $z=1$ to $z=-e$, the argument increases continuously from $\theta=0$ to $\theta=\pi$ (approaching $-e=e\cdot e^{i\pi}$ from above):
$$ \boxed{f^*(-e)=\sqrt e\,e^{i\pi/2}=i\sqrt e} $$
Part $2^\circ$: path through the lower half-plane
Moving through the lower half-plane instead, the argument decreases continuously from $\theta=0$ to $\theta=-\pi$ (approaching $-e=e\cdot e^{-i\pi}$ from below):
$$ \boxed{f^*(-e)=\sqrt e\,e^{-i\pi/2}=-i\sqrt e} $$
Why the two paths disagree
The two paths differ by one full loop around the branch point $z=0$: going upper-then-lower (or vice versa) back to the same target point traces a closed loop encircling $0$ once, and $\sqrt z$ picks up a factor of $e^{i\pi}=-1$ per loop. Indeed:
$$ \frac{i\sqrt e}{-i\sqrt e}=-1 $$
exactly the sign flip expected after one circuit around a simple square-root branch point.
Branch-cut viewpoint
With the standard cut along the negative real axis, approaching $z=-e$ from above corresponds to $\arg z\to\pi^-$, and from below to $\arg z\to-\pi^+$ — the two boundary values of the cut, which is exactly why they differ.
Result
A direct illustration of why $\sqrt z$ needs a branch cut: the "value at $-e$" is genuinely path-dependent unless one fixes which side of the cut is being approached.