Let's parameterize the curve $\gamma$ as $z=1+e^{it}$, where $t$ starts at $\pi/2$ and ends at $-\pi/2$. Then,
$$\begin{align}
\int_{\gamma}\sqrt{z-1}\,dz&=\int_{\pi/2}^{-\pi/2}\sqrt{e^{it}}ie^{it}\,dt\\\\
&=\frac23 \left.\left(e^{it}\right)^{3/2}\right|_{\pi/2}^{-\pi/2}\\\\
&=-i\frac{2\sqrt{2}}{3}
\end{align}$$
To show that the imaginary part of f(z) is not multivalued for any branch of the logarithm, we need to ensure that the argument of the logarithm is continuous as we vary z in the complex plane.
Given the function $f(z)$ defined as
\begin{equation}
f(z) = \int_{\gamma} \frac{dw}{w - z},
\end{equation}
where $\gamma(t) = t$, $0 < t < 1$, and $z \notin [0, 1]$.
Let's compute the integral:
\begin{equation}
f(z) = \int_{0}^{1} \frac{dt}{t - z} = \ln(|1 - z|) - \ln(|z|) + i(\arg(1 - z) - \arg(-z)).
\end{equation}
Since $z \notin [0, 1]$, there is no risk of the real parts of the logarithms exploding.
To ensure the imaginary part is not multivalued for any branch of the logarithm, we must show that $\arg(1 - z)$ and $\arg(-z)$ are continuous as we vary $z$ in the complex plane. This is guaranteed as long as we avoid the branch cut.
The branch cut for $\arg$ typically lies along the negative real axis. Therefore, to ensure continuity, avoid any path for $z$ that crosses the negative real axis.
When $z \notin [0, 1]$, $f(z)$ is well defined, and the imaginary part is not multivalued for any branch of the logarithm, provided we avoid crossing the negative real axis when varying $z$ in the complex plane.
====================REPLY TO Conrad================
He/She is correct; the argument I provided addresses the general case where z is complex and does not specifically cover the case where z is a real positive number greater than 1.
When z is a real positive number greater than 1, the function f(z) is still well-defined, and you can use the same argument regarding the continuity of the imaginary part. Specifically, $\arg(1 - z)$ and $\arg(-z)$ remain continuous as long as you avoid crossing the branch cut of the argument function, which is typically along the negative real axis.
Since z is real and greater than 1, it does not cross the branch cut, and there are no issues with the continuity of the imaginary part. Therefore, the function f(z) is well-defined for z as a real positive number greater than 1.
Best Answer
The key is: from the definition $\log(z)=\log|z|+i\arg z$ where $\arg z\in [-\pi,\pi)$, it is clear that $\overline{\log z}=\log \bar z$ on $\mathbb C\setminus \mathbb R_{\le 0}$.
Hence, by parametrizing the path as $z=\gamma(t)=-i+e^{i\pi/4}t$, $t\in[0,\sqrt 2]$, $$\int_{\gamma}\overline{\operatorname{Log}(z)}dz=\int_0^{\sqrt 2}\overline{\operatorname{Log}(-i+e^{i\pi/4}t)}e^{i\pi/4}dt=e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt$$
At this point, we have two approaches:
Direct calculation $$\begin{align} e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt &=e^{i\pi/4}\cdot\frac{(i+e^{-i\pi/4}t)(\operatorname{Log}(i+e^{-i\pi/4}t)-1)}{e^{-i\pi/4}}\bigg\vert^{t=\sqrt 2}_{t=0} \\ &= -1+\left(\frac{\pi}{2}-1\right)i \end{align} $$
Complex analysis approach $$\begin{align} e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt &=i\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)e^{-i\pi/4}dt\\ &= i\int_{\bar\gamma}\operatorname{Log}(z)dz \quad(1)\\ &= -i\int_{\text{quarter arc}}\operatorname{Log}(z)dz \quad(2)\\ &= -i\int^{\pi/2}_0(\ln|e^{i\theta}|+i\theta)ie^{i\theta}d\theta \\ &= i\int^{\pi/2}_0\theta e^{i\theta}d\theta \\ &=-1+\left(\frac{\pi}{2}-1\right)i \end{align} $$
$(1)$: $\bar\gamma$ is the straight line from $+i$ to $1$.
$(2)$: $\text{quarter arc}$ is the quarter arc from $+i$ to $1$. The deformation of integration path is justified by Cauchy's integral theorem.
Although method 1 looks simpler, indeed the algebra is more tedious. As a complex analysis enthusiaist, I view method 2 as a more elegant solution.