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.
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.)
My parametrization was the following: \begin{alignat}{5} \gamma_1 &:& z &=& 3 + i(2t-1), &&\quad 0\leq t\leq 1\\ \gamma_2 &:& z &=& 9 - 6t + i, &&{} 1\leq t\leq 2\\ \gamma_3 &:& z &=& -3 + i(5 - 2t), &&{} 2\leq t\leq 3\\ \gamma_4 &:& z &=& -21 + 6t - i, &&{} 3\leq t\leq 4 \end{alignat} Then my integral is $$ \int_0^1\frac{3 - i(2t-1)}{11 + i(2t - 1)}(2idt) + \int_1^2\frac{9 - 6t - i}{17 - 6t + i}(-6dt) + \int_2^3\frac{-3 - i(5 - 2t)}{5 + i(5 - 2t)}(-2idt) + \int_3^4\frac{-21 + 6t + i}{-13 + 6t - i}(6dt) $$ Now multiplying through by the conjugate, we have \begin{align} \int_0^1\frac{3 - i(2t-1)}{11 + i(2t - 1)}(2idt) &= 2i\int_0^1\frac{33-(2t-1)^2}{121+(2t-1)^2}dt - 28\int_0^1\frac{1-2t}{121+(2t-1)^2}dt\\ &= 2i\Big(-1+14\tan^{-1}\Big(\frac{1}{11}\Big)\Big)\\ \int_1^2\frac{9 - 6t - i}{17 - 6t + i}(-6dt) &= -6\int_0^1\frac{152-156t+36t^2}{(17 - 6t)^2+1}dt - 12i\int_0^1\frac{6t-13}{(17 - 6t)^2+1}dt\\ &= -6 + (4+i)\ln\Big(\frac{775-168i}{169}\Big)\\ \int_2^3\frac{-3 - i(5 - 2t)}{5 + i(5 - 2t)}(-2idt) &= -2i\int_2^3\frac{-40-4t^2+20t}{25+(5-2t)^2}dt + 4\int_2^3\frac{2t-5}{25+(5-2t)^2}dt\\ &= -2i\Big[2\tan^{-1}\Big(\frac{1}{5}\Big) - 1\Big]\\ \int_3^4\frac{-21 + 6t + i}{-13 + 6t - i}(6dt) &= 6\int_3^4\frac{36t^2-204t+272}{(-13 + 6t)^2 + 1}dt + 12i\int_3^4\frac{6t-17}{(-13 + 6t)^2 + 1}dt\\ &= 6-(1+4i)\tan^{-1}\Big(\frac{168}{775}\Big)-(8-2i)\tanh^{-1}\Big(\frac{24}{37}\Big) \end{align} Hopefully I dont have an small errors since that was tedious. The integral is then equal to $$ 3.13297i $$
