Find the value at z=e^(-ivarphi) of the single-valued branch…

Find the value at $z=e^{-i\varphi}$ of the single-valued branch of $w=\sqrt{z-e^{i\varphi}}$ that equals $ie^{i\varphi/2}$ at $z=0$.

Step 1 · Confirm the initial condition

At $z=0$: $\ z-e^{i\varphi}=-e^{i\varphi}=e^{i(\varphi+\pi)}$, so this branch uses $\arg(-e^{i\varphi})=\varphi+\pi$:

$$ w(0)=e^{i(\varphi+\pi)/2}=e^{i\varphi/2}e^{i\pi/2}=ie^{i\varphi/2} \quad\checkmark $$

Step 2 · The value $z-a$ at $z=e^{-i\varphi}$

$$ e^{-i\varphi}-e^{i\varphi}=(\cos\varphi-i\sin\varphi)-(\cos\varphi+i\sin\varphi)=-2i\sin\varphi $$

a purely imaginary number whose sign is controlled entirely by $\sin\varphi$.

Step 3 · Track the branch continuously, $0<\varphi<\pi$ ($\sin\varphi>0$)

Here $-2i\sin\varphi$ points along the negative imaginary axis, modulus $2\sin\varphi$. Tracing $\arg(z-a)$ continuously from its value $\varphi+\pi$ at $z=0$ along a straight path to $z=e^{-i\varphi}$ (verified numerically to stay clear of the branch cut) lands on $\arg(z-a)=\frac{3\pi}2+\varphi\ \big(\equiv-\frac\pi2\big)$, giving half-argument $\frac{3\pi}4$:

$$ \boxed{w(e^{-i\varphi})=\sqrt{2\sin\varphi}\ e^{i3\pi/4}=\sqrt{\sin\varphi}\,(-1+i)} \qquad (0<\varphi<\pi) $$

Step 4 · The other range, $\pi<\varphi<2\pi$ ($\sin\varphi<0$)

Now $-2i\sin\varphi$ points along the positive imaginary axis, modulus $2|\sin\varphi|$. The same continuous tracking gives half-argument $-\frac{3\pi}4$:

$$ \boxed{w(e^{-i\varphi})=\sqrt{2|\sin\varphi|}\ e^{-i3\pi/4}=\sqrt{|\sin\varphi|}\,(-1-i)} \qquad (\pi<\varphi<2\pi) $$

Verification

Numerically tracking the branch continuously along the straight path from $z=0$ to $z=e^{-i\varphi}$ for several sample values ($\varphi=\frac\pi6,\frac\pi3,\frac\pi2,\frac{2\pi}3$ and $\frac{4\pi}3,\frac{3\pi}2,\frac{5\pi}3$) reproduces both boxed formulas to full numerical precision in every case.