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I am having trouble computing an integral along a line segment on complex plane. Namely:

Let $\operatorname{Log}{z}$ denote the principal branch of logarithm on $\mathbb{C}-\mathbb{R_{\leq0}}$ and let $\gamma$ be a line segment from $-i$ to $1$. Compute the integral: $$\int_{\gamma}\overline{\operatorname{Log}(z)}dz$$


Attempt: The function $\overline{\operatorname{Log}(z)}$ is not holomorphic so we are not sure whether an antiderivative exists, so what seemed resonable for me was to parametrize $\gamma$ in polar coordinates and try direct computation by $\int_{\gamma}\log(|z|) + i\int_{\gamma}Arg(z)$ but the parametrization is rather ugly which makes computation rather too difficult taking into account that this was an exam task. Is there any neat way to compute this?

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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:

  1. Direct calculation $$\begin{align} e^{i\pi/4}\int_0^{\sqrt 2}\operatorname{Log}(i+e^{-i\pi/4}t)dt &=e^{i\pi/4}\cdot\frac{(i+e^{-i\pi/4}t)(\operatorname{Log}(i+e^{-i\pi/4}t)-1)}{e^{-i\pi/4}}\bigg\vert^{t=\sqrt 2}_{t=0} \\ &= -1+\left(\frac{\pi}{2}-1\right)i \end{align} $$

  2. 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.

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