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In my complex analysis book, there is an example where I am asked to compute $\int_\Gamma1/z \, dz$ for two cases: in both of them, $\Gamma$ is a curve going from $-i$ to $i$ in the complex plane. However, in the first case, $\Gamma$ lies in both the first and fourth quadrants, crossing the positive real axis. Let us call it $\Gamma_1$. In the second case, $\Gamma$ lies in both the second and third quadrants, crossing the negative real axis. Let us call it $\Gamma_2$.

Computing

$\int_{\Gamma_1}\frac{dz}{z}=\log z|_{-i}^i =\pi i$

is simple enough. However, computing

$\int_{\Gamma_2}\frac{dz}{z}$

requires that I make some changes due to a "branch cut" issue with the $\log$ function. The book goes off to explain that, in this case, I have to use $\log|z|+i\arg z$.

I am very confused: where did that come from? Thanks in advance for your help!

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    You can just choose a different logarithm and put the branch cut somewhere else. OR you could just calculate $\int_{\Gamma_1-\Gamma_2} dz/z$.2012-06-25

1 Answers 1

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Why not apply directly the definition of line integral?

As you don't give the actual paths I assume any path from $\,-i\,$ to $\,i\,$ passing through quadrants $\,II-III\,$ is fine, so let us take the path $\gamma:=\left\{z\in\Bbb C\;:\;|z|=1\,,\,-\frac{\pi}{2}\leq\arg z\leq\frac{\pi}{2}\right\}$ and thus on this path we have that $z=e^{it}\,,\,-\frac{\pi}{2}\leq t \leq\frac{\pi}{2}\Longrightarrow dz=ie^{it}dt$ so that $\int_\gamma\frac{dz}{z}=i\int_{-\pi/2}^{\pi/2}dt=\pi i$ and you don't need to mess with any branch cuts of anything at all.

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    Thank you for clarifying this to me. :)2012-09-06