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I'm working on some past exam papers and I wanted to see if I'm thinking of the following in the correct way;

Given a contour integral $\int_{\gamma}\frac{1}{z}dz$ I want to find all of it's possible values given that $\gamma$ is the semi-circle starting at $1$ and going to $-1$

My thinking is that as the radius of the semicircle is $1$ then $|z|=1$ and so $z=e^{i\theta}$.

We can also parametrise $\gamma$ as $e^{i\theta}, \theta \in [-\pi,\pi]$

This means we can express our integral as;

$\int_{\gamma}\frac{1}{z}dz=\int_{\pi}^{-\pi}e^{-i\theta}ie^{i\theta}d\theta=\int_{\pi}^{-\pi}id\theta=0$

So what does the question mean by find all the values of $\int_{\gamma}\frac{1}{z}$ ?

It seems to me there is only one ?

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    $\begingroup$ There are two different arcs you can take. Also you calculated the integral wrong and your parameterization is wrong. $\endgroup$
    – Cameron Williams
    May 10, 2018 at 22:58
  • $\begingroup$ @CameronWilliams do you mean I should evaluate both $\int_{\pi}^{-\pi}id\theta$ and $\int_{-\pi}^{\pi}id\theta$ ? wouldn't they both still just give zero ? or should I evaluate $\int_{\pi}^{-\pi}e^{-i\theta}ie^{i\theta}d\theta$ and $\int_{\pi}^{-\pi}e^{-i\theta}ie^{-i\theta}d\theta$ ? $\endgroup$
    – excalibirr
    May 10, 2018 at 23:02
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    $\begingroup$ @exodius The point (amongst other points) is $\int_{\pi}^{-\pi} d\theta \ne 0.$ It is $-2\pi.$ $\endgroup$
    – spaceisdarkgreen
    May 10, 2018 at 23:03
  • $\begingroup$ @spaceisdarkgreen oh shoot sorry , I didn't see my mistake until now , So then I should have the integrals $\int_{\pi}^{-\pi}id\theta=2\pi i$ and $\int_{-\pi}^{\pi}id\theta=-2\pi i$ ? $\endgroup$
    – excalibirr
    May 10, 2018 at 23:05
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    $\begingroup$ @exodius It really depends whether you're taking the upper or lower semicircle. If you are taking the upper semicircle, from $1$ to $-1,$ then $e^{i\theta}$ for $\theta\in[0,\pi]$ is the most natural parametrization. You then get $\int_{0}^\pi e^{-i\theta}ie^{i\theta}d\theta = i\pi.$ ($-\pi$ and $\pi$ are the same angle, so going between these would be going around a whole circle, starting and ending at $-1$). $\endgroup$
    – spaceisdarkgreen
    May 10, 2018 at 23:08

1 Answer 1

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Given a Contour integral $\int_{\gamma}\frac{1}{z}dz$ where $\gamma$ is the semicircle going from $1$ to $-1$, then there two arcs which we can integrate over.

There is the lower semi-circle, where we go from $1$ to $-1$ in a clockwise direction, and also the upper semi-circle , where we go from $1$ to $-1$ in an anti-clockwise direction .

With that in mind there are two semi-circles we can choose from;

1) moving in a clockwise direction in which case $\gamma$ can be parametrised as $e^{-i\theta}, \theta \in[0,\pi]$

So here we have $\int_{\gamma}\frac{1}{z}dz=\int_{0}^{\pi}\frac{-ie^{-i\theta }d\theta}{e^{-i\theta}}$, as z=$e^{-i\theta}$ when moving counter-clockwise

$\int_{0}^{\pi}\frac{-ie^{-i\theta }d\theta}{e^{-i\theta}}=-i\int_{0}^{\pi}d\theta=-i[\theta]^{\pi}_{0}=-i\pi$

2) moving in a counter clockwise direction , in which case we can parametrise $\gamma$ as $e^{i\theta}, \theta \in [0,\pi]$

So for this we have $\int_{\gamma}\frac{1}{z}dz=\int_{0}^{\pi}\frac{ie^{i\theta }d\theta}{e^{i\theta}}$, as z=$e^{i\theta}$

$\int_{0}^{\pi}\frac{ie^{i\theta }d\theta}{e^{i\theta}}=i\int_{0}^{\pi}d\theta=i[\theta]^{\pi}_0=\pi i$

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