f^*(z) is the branch of f(z)=√(z-1) for which f^*(0)=-1 .…

$f^*(z)$ is the branch of $f(z)=\sqrt{z-1}$ for which $f^*(0)=-1$. Find $f^*(-i)$.

Step 1 · The stated condition is inconsistent

At $z=0$: $\ z-1=-1$, so $f^*(0)=\sqrt{-1}$, which can only equal $i$ or $-i$ — never $-1$ (whose square is $1$, not $-1$). The condition "$f^*(0)=-1$" as literally stated is impossible for any branch of $\sqrt{z-1}$; this is evidently a typo for $f^*(0)=-i$, which we adopt below.

Step 2 · Fix the branch using $f^*(0)=-i$

Since $z-1=-1$ at $z=0$, and $\sqrt{-1}=-i$ requires $\arg(-1)=-\pi$ on this branch (so that half-argument $=-\pi/2$ gives $e^{-i\pi/2}=-i$), this branch uses:

$$ \arg(-1)=-\pi \quad\text{(rather than the principal value $+\pi$)} $$

Step 3 · Evaluate $z-1$ at $z=-i$

$$ z-1=-i-1=-1-i, \qquad |z-1|=\sqrt2, \qquad \text{principal }\arg(-1-i)=-\frac{3\pi}4 $$

(third quadrant, both parts negative). A straight path from $0$ to $-i$ stays clear of the branch point $z=1$ and reaches this value while continuously tracking the branch fixed in Step 2.

Step 4 · Compute $f^*(-i)$

$$ f^*(-i)=|{-1-i}|^{1/2}\exp\left(\frac i2\cdot\left(-\frac{3\pi}4\right)\right)=2^{1/4}e^{-3\pi i/8} $$

$$ =2^{1/4}\left[\cos\frac{3\pi}8-i\sin\frac{3\pi}8\right]\approx0.4551-1.0987i $$

Verification

Numerically tracking the branch (choosing at each step along the path the square root closer to the previous value) from $z=0$ with $f^*(0)=-i$ to $z=-i$ reproduces $0.45509\ldots-1.09868\ldots i$ to full precision, matching $2^{1/4}e^{-3\pi i/8}$ exactly.