8
$\begingroup$

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.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Have you tried parameterizing $\gamma$ like you would with any ordinary line integral? i.e., break up $\gamma$ into four sub-countours, call them $\gamma_1,\gamma_2,\gamma_3,\gamma_4$, then compute each integral. $\endgroup$ – user170231 Sep 22 '14 at 22:01
  • $\begingroup$ Yes...I used the standard parameterization of the form $tp_1+(1-t)p_2$ but the integrals are leading to answers strange enough that I am not sure I am doing it correctly. $\endgroup$ – Johnny Apple Sep 23 '14 at 1:52
  • $\begingroup$ Just so it's said here: this integral cannot be evaluated using residues, as the integrand is not analytic inside the rectangle (or anywhere else). $\endgroup$ – Ron Gordon Oct 27 '14 at 13:38
9
+50
$\begingroup$

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.)

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Concise and smart! One aspect we also learn from (1) is to simplify the work by splitting the integrand into a holomorphic part which vanishes when integrating along a closed curve and in a non-analytic part. (+1) $\endgroup$ – Markus Scheuer Oct 27 '14 at 14:17
5
$\begingroup$

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 $$ 3.13297i $$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Looking at your $\gamma_k$ it seems that you computed the counterclockwise integral; but the OP wanted it clockwise. $\endgroup$ – Christian Blatter Oct 27 '14 at 19:06
  • $\begingroup$ @ChristianBlatter oh I misread $\endgroup$ – dustin Oct 27 '14 at 20:18
  • $\begingroup$ Thank you dustin. This was the approach that I had, but I assumed that I goofed somewhere since it got a bit messy. $\endgroup$ – The Substitute Oct 28 '14 at 0:54
  • $\begingroup$ @TheSubstitute no problem but note I went the wrong direction around just reverse to get the correct solution which picks up a factor of negative 1 $\endgroup$ – dustin Oct 28 '14 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.