Simple complex line integral over a rectangle What is the easiest way without using residues to calculate:
$$\int_{\gamma} {\overline z \over {8 + z}} dz$$
Where $\gamma$ is the rectangle with vertices $\pm 3 \pm i$ in $\Bbb C$ in the clockwise direction.
Am I destined to do some long tedious substitution process here? All I know so far is Cauchy Integral formula, Cauchy's theorem, definition of line integral and various theorems related to it, including a sort of Fundamental Theorem for complex line integrals.
 A: Some preliminaries: For $a$, $b\in{\mathbb C}$ denote by $[a,b]$ the segment path beginning at $a$ and ending at $b$. When $a$ and $b$ have absolute value $<8$ then ${\rm Re}(8+z)>0$ along $[a,b]$, and therefore
$$\int_{[a,b]}{z+c\over 8+z}\ dz=\bigl(z+(c-8){\rm Log}(8+z)\bigr)\biggr|_a^b\tag{1}$$
whatever $c\in{\mathbb C}$.
Put $\gamma_k:=[z_{k-1},z_k]$ $(1\leq k\leq4)$, where the $z_k$ are the vertices of the rectangle in clockwise order, with $z_0=z_4=-3-i$.
On $\gamma_1$ one has $\bar z=-(z+6)$, on $\gamma_2$ one has $\bar z=z-2i$, then on $\gamma_3$ one has $\bar z=-(z-6)$, and finally on $\gamma_4$ one has $\bar z=z+2i$. It follows that
$$I:=\int_\gamma{\bar z\over 8+z}\ dz=-\int_{\gamma_1}{z+6\over 8+z}\ dz+\int_{\gamma_2}{z-2i\over 8+z}\ dz-\int_{\gamma_3}{z-6\over 8+z}\ dz+\int_{\gamma_4}{z+2i\over 8+z}\ dz\ .$$
The integrals appearing on the right can be evaluated by means of $(1)$:
$$\eqalign{-\int_{\gamma_1}&=z_0-z_1+2{\rm Log}(8+z_1)-2{\rm Log}(8+z_0),\cr
\int_{\gamma_2}&=z_2-z_1-(8+2i){\rm Log}(8+z_2)+(8+2i){\rm Log}(8+z_1),\cr
-\int_{\gamma_3}&=z_2-z_3+14{\rm Log}(8+z_3)-14{\rm Log}(8+z_2),\cr
\int_{\gamma_4}&=z_0-z_3-(8-2i){\rm Log}(8+z_0)+(8-2i){\rm Log}(8+z_3)\ .\cr}$$
Summing it all up gives
$$\eqalign{I&=2(z_0-z_1+z_2-z_3)\cr
&\quad-(10-2i){\rm Log}(5-i)+(10+2i){\rm Log}(5+i)\cr &\quad-(22+2i){\rm Log}(11+i)+(22-2i){\rm Log}(11-i)\cr
&=2i\>{\rm Im}\bigl((10+2i){\rm Log}(5+i)-(22+2i){\rm Log}(11+i)\bigr),\cr}$$
since the alternating sum of the $z_k$ vanishes, and $w-\bar w=2i\>{\rm Im}(w)$. When $p>0$ then $${\rm Log}(p+ i)={1\over2}\log(p^2+1)+i \arctan{1\over p}\ .$$
Therefore we finally obtain
$$I=2i\left(\log 26+10\arctan{1\over5}-\log 122-22\arctan{1\over11}\right)\doteq-3.13297\>i\ .$$
(That ${\rm Re}(I)=0$ could have been detected in advance using symmetry considerations.)
