Closed path integral

Question

Thanks for coming in

This is a very basic question and I'm sorry to ask about it, since I really couldn't figure it out

I have searched for closed path integral on this site and found a lot of them but still couldn't figure it out why? let say I have closed path $S$ from $(1,-i)\space\space\space\rightarrow(1,i)\\(1,i)\space\space\space\space\space\space\rightarrow(-1,i)\\(-1,i)\space\space\space\rightarrow(-1,-i)\\(-1,-i)\rightarrow(1,-i)$

and $\int_S\frac{2z+6}{z^2+6z+13}$dz

First, I did as following

$\int\frac{2z+6}{z^2+6z+13}dz=\ln(z^2+6z+13)$ for each path let says $S_1,S_2,S_3,S_4$and then I summed them together

which gave me zero as an answer

however, I also found that if I did it differently, it could give me a different answer (a correct one, I believe) by

$\int_S\frac{2z+6}{z^2+6z+13}dz = \int_S \frac{1}{z+3-2i}dz+\int_S\frac{1}{z+3+2i}dz$

(edit#1begin)

$\int_S \frac{1}{z+3-2i} = \int_{1-i}^{1+i}\frac{1}{z+3-2i}+\int_{1+i}^{-1+i}\frac{1}{z+3-2i}+\int_{-1+i}^{-1-i}\frac{1}{z+3-2i}+\int_{-1-i}^{1-i}\frac{1}{z+3-2i}\\ = \ln\frac{4-i}{4-3i}\frac{2-i}{4-i}\frac{2-3i}{2-i}\frac{4-3i}{2-3i} = \ln(1)\\ = \ln(e^{2\pi i}) = 2\pi i $

(edit#1end)

then I got an answer as $4\pi i$ instead

So, my question is why can't I go with the first solution? what did I miss?

Thank you

Answer 1

How about this as a simpler example. Compute $$\int_S\frac{dz}{z}$$ over your contour. The answer is $2\pi i$ by calculus of residues. But you could say the indefinite integral is $\ln(z)$ and sum as you did over each edge and get zero.

This is fallacious of course: $\ln$ is a "multivalued" function on the nonzero complexes, and there is no branch of the function that is continuous on the whole contour $S$. The same problem holds in your example. There is no continuous branch of $\ln(z^2+6z+13)$ defined on $S$. By the time you've gone round the contour you've hit a different value of $\ln(z^2+6z+13)$ to that you started with. These different values must differ by a multiple of $2\pi i$, so an answer of $4\pi i$ is quite plausible...

Answer 2

The solutions of the equation $z^2+6z+13=0$ are $-3\pm 2i$. Hence there is a region $D$ such that $S \subseteq D$, $-3\pm 2i \notin D$ and $f(z):=\frac{2z+6}{z^2+6z+13}$ is holomorphic on $D$. Then Cauchy says:

$$\int_Sf(z) dz =0.$$